Class 11 Medical Physics - Motion In A Plane

A ball is thrown from the top of a tower with an initial velocity of 10 $ms^{-1}$ at an angle of $30^0$ above the horizontal. It hits the ground at a distance of 17.3 m from the base of the tower. Calculate the height of the tower. (g = 10 $ms^{-2}$ )

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A ball is thrown upwards from the top of a tower 40 m high with a velocity of 10 m/s. Find the time when it strikes the ground (Take g = 10 $m/s^{2}$ ).

Taking upward direction as

$+$

and the downward direction as

$-$

with using

$s=ut+\dfrac{1}{2}at^2$

we need to put

$u=10m/s$

$a=-g=-10m/s^2$ and

$s=-40m$

$\text{(downward *displacement*)}$

we get the following equation,

$-40=10t-5t^2$

$t^2-2t-8=0$ or

$t=4second$

Just employ the Shridharacharya Formula

$t=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

Where

$t$ is root of

$at^2+bt+c=0$

A particle is projected from ground at some angle to the horizontal. $($ Assuming point of projection to be origin and the horizontal and vertical directions to be the $x\ \&\ y$ axis $)$ the particle passes through the points $(3, 4)$ and $(4, 3)$ during its motion. If the range of the particle is $\dfrac{37}{x}\ m$ then $x$ is:

The equation of trajectory:

$y=x\tan \alpha \left ( 1-\dfrac{x}{R} \right )$

Substituting the point (3,4)

$4=3\tan \alpha \left ( 1-\dfrac{3}{R} \right ) ------- \left ( i \right )$

Substituting the point (4,3)

$3=4\tan \alpha \left ( 1-\dfrac{4}{R} \right ) ------- \left ( ii \right )$

Dividing

$\left ( i \right )$ and

$\left ( ii \right )$ , we get :

$\dfrac{4}{3} =\dfrac{3}{4}\dfrac{1-\dfrac{3}{R}}{1-\dfrac{4}{R}}$

$\Rightarrow 16\left ( 1-\dfrac{4}{R} \right ) =9\left ( 1-\dfrac{3}{R} \right )$

$\Rightarrow 7 = \dfrac{64}{R} - \dfrac{27}{R}$

$\Rightarrow R = \dfrac{37}{7}$

The resultant of two like parallel forces 12 N and 7 N is ________.

The resultant of two like parallel forces

$=$ addition of the two forces

$=12+7$

$=19N$

$\displaystyle \bigtriangleup ABC$ is right-angled at C If AC = 5 cm and BC=12 cm find the length of AB

Vector Addition

REF.Image

Given ,

$\Delta ABC$ is right angled at C.

$AC=5 cm$

$BC=12 cm$

$AB=?$

Using Pythagoras theorem.

REF.Image

$c^{2}=a^{2}+b^{2}$

$c=\sqrt{a^{2}+b^{2}}$

so in

$\Delta ABC$

$AB=\sqrt{(AC)^{2}+(CB)^{2}}$

$AB=\sqrt{(5)^{2}+(12)^{2}}$

$AB=\sqrt{25+144}$

$AB=\sqrt{169}$

$AB=13 cm$

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A car is moving with speed $30\ m/sec$ on a circular path of radius $500\ m$ . Its speed is increasing at the rate of $2\ m/sec^{2}$ . What is the net acceleration of the car at that moment (i.e the vector sum of tangential and radial acceleration)?

Velocity,

$v=30$

${m/s}$

Radius,

$r=500$

$m$

Net acceleration is given by:

$\overrightarrow a=(\ddot r-r\dot \theta^2)\hat r+(r\ddot \theta+2\dot r\dot \theta)\hat \theta$

where,

$\hat r$ is radial direction,

$\hat \theta$ is tangential direction.

$\dot r=\ddot r=0$ , Since, the radial distance is not changing. Also,

$\ddot \theta=2$

${m/s^2}$

$\therefore \overrightarrow a=\left(0-500\times \left(\cfrac{V}{r}\right)^2\right)\hat r+(500\times 2+0)\hat \theta$

$\implies \overrightarrow a=-500\times \cfrac{900}{250000}\hat r+1000\hat \theta$

$\implies \overrightarrow a=-1.8\hat r+1000\hat \theta$

$|\overrightarrow a|=\sqrt{(1.8)^2+(1000)^2}$

$\implies |\overrightarrow a|=\sqrt{3.24+1000000}$

$\implies |\overrightarrow a|=1000.0017$

${m/s^2}$

Two forces $5$ kg wt and $10$ kg wt are acting with an inclination of $120^o$ between them. what is the angle the resultant makes with $10$ kg wt?

Let P be the force of 5 kg weight and Q be the force of 10 kg weight.

By parallelogram law of vector addition, the resultant is,

${{\rm{R}}^2} = {{\rm{P}}^2} + {{\rm{Q}}^2} + 2{\rm{PQ}}\cos \theta$

${{\rm{R}}^2} = 25 + 100 + 100\cos 120^\circ$

${\rm{R}} = 5\sqrt 3 \;{\rm{weight}}$

Let $\alpha$ be the angle made by R with 10 kg weight.

$\tan \alpha = \frac{{{\rm{P}}\sin \theta }}{{{\rm{P}} + {\rm{Q}}\cos \theta }}$

$\tan \alpha = \frac{{5\sin 120}}{{5 + 10\cos 120}}$

$\tan \alpha = \frac{1}{{\sqrt 3 }}$

$\alpha = 30^\circ$

Three point charges $+Q, + 2Q$ and $-3Q$ are placed at the vertices of an equilateral triangle ABC of side L.if these charges are displaced to the mid-points $A_1, B_1$ and $C_1$ respectively find the amount of the work done in shifting the charges to the new locations.

The charges are : - +q, +2Q and +Q and the potential energy is zero. Let the side of triangle be r

Equating P.E of system to zero

$k/r [ ( q x 2Q) + (2Q x Q) + (q x Q) ] = 0$

$2Q + 3q = 0$

$Q = (-3q)/2$

Find a unit vector parallel to the resultant of the vectors $\bar{A}=2\hat{i}+3\hat{J}+4\hat{k}$ and $\bar{B}=3\hat{i}-5\hat{J}$ .

$\textbf{Step 1: Calculate Resultant of two vectors}$

$\vec{R} = \vec{A} + \vec{B}$

$\textbf{Step 2: Calculate unit vector parallel to}$

$\vec{R}$

$\hat{R} = \dfrac{\vec{R}}{|\vec{R}|}$

$\Rightarrow$

$\hat{R} = \dfrac{\vec{A} + \vec{B}}{|\vec{A} + \vec{B}|}$

$-------------------(1)$

$\textbf{Step 3:}$

$\textbf{Calculations}$

$\vec{A} + \vec{B} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + (3 \hat{i} - 5 \hat{j})$

$\Rightarrow$

$\vec{A} + \vec{B} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$

$\Rightarrow$

$|\vec{A} + \vec{B}| = \sqrt{25 + 4 + 16}$

$= \sqrt{45}$

From equation

$(1)$

$\therefore$

$\hat{R} = \dfrac{5 \hat{i} - 2 \hat{j} + 4 \hat{k}}{\sqrt{45}}$

Unit vector parellel to resultant will be same as unit vector along resultant (because direction will be same)

A ball is thrown from the top of a building 45 m high with a speed 20 m s $^{-1}$ above the horizontal at an angle of $30^o$ . Find
a. The time taken by the ball to reach the ground.
b. The speed of ball just before it touches the ground.

Given

$v = 20 \,m s^{-1}, \theta= 30^o, H= 45\, m$ .

a. As the ball has been projected at an angle of

$30^o$ above horizontal, so first of all we need to analyse the velocity horizontally and vertically. This will be useful while using distance-time relation in horizontal and vertical directions.

$\displaystyle v_{xi} = v\,cos\,30^o = 20 \times \frac{\sqrt 3}{2} = 10 \sqrt 3 \, m\,s^{-1}$

$v_{yi} =v\,sin \,30^o = 20 \times \frac{1}{2} = 10 \,m\,s^{-1}$

It will be easy for us to use distance-time relation in vertical as it will involve less calculation.

In y-direction:

$\displaystyle -45 = 10t -\frac{1}{2} \times gt^2 \,\Rightarrow \,t^2 - 2t = 0$

which on solving gives

$t = 1 +\sqrt 10\,s$ (positive value), (other value is

$1 - \sqrt 10\,s$ , a negative value of time is not acceptable).

$v_{yf} = 10 - 10 \times (1 + \sqrt 10) = -10 \sqrt 10 \,m\,s^{-1}$

$v_f = \sqrt { v^2_{yf} + v^2_{xf} }$

$= \sqrt {(10\sqrt 10)^2 + (10 \sqrt 3)^2 }$

$= 10 \sqrt 3\,m\,s^{-1}$

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Two bullets are fired with horizontal velocities of $50m/s$ and $100m/s$ from two guns at a height of $19.6m$ (a) Will both the bullets strike the ground? (b) If yes, then after how much time and which bullet will strike first? (c) What would be the path of the bullets? ( $g=9.8m/{s}^{2}$ )

$(a)$ Yes, both the bullets will strike the ground at the same time.because the downward acceleration and initial velocity in the downward direction of the two bullets are same. They will take the same time to hit the ground and for half projectile.

$(b)$ Height

$=19.6m$

$v=0$

$h = vt + \dfrac{1}{2}g{t^2}$

$\begin{array}{l} \Rightarrow 19.6=0\times t+\dfrac { 1 }{ 2 } \times 9.8\times { t^{ 2 } } \\ \Rightarrow { t^{ 2 } }=\dfrac { { 19.6 } }{ { 4.9 } } =4 \\ t=2\sec \end{array}$

Hence, both bullets will hit the ground simultaneously after

$2{\rm{ }}second$ ,at the same.

c)Both the bullets travel half projectile, the bullet with higher velocity cover more distance then the other velocity.

Two forces each of $10\ N$ act at an angle ${60}^{o}$ with each other. The magnitude & direction of the resultant with respect to one of the vectors is

From the formula

$\left| \overrightarrow { C } \right| =\sqrt { { A }^{ 2 }+{ B }^{ 2 }+2AB\cos { \theta } } \\ \quad \quad \quad =\sqrt { 100+100+2\times 10\times 10\times \cfrac { 1 }{ 2 } } \\ \quad \quad \quad =\sqrt { 300 }$

Magnitude

$=10\sqrt { 3 }$

Direction :

$\angle \phi ={ tan }^{ -1 }\left( \cfrac { B\sin { \theta } }{ A+B\cos { \theta } } \right) \\ \quad \quad \quad ={ tan }^{ -1 }\left( \frac { 10\times \cfrac { \sqrt { 3 } }{ 2 } }{ 10+10\times \cfrac { 1 }{ 2 } } \right) \\ \quad \quad \quad ={ tan }^{ -1 }\left( \cfrac { 5\sqrt { 3 } }{ 15 } \right) \\ \quad \quad \quad ={ tan }^{ -1 }\left( \cfrac { \sqrt { 3 } }{ 3 } \right)$

A projectile fired from the top of a $40\ m$ high cliff with an initial speed of $50\ m/s$ at an unknown angle. Find its speed when it hits the ground. $(g=10\ {m/s}^{2})$

by conservation of energy

$mg40+\dfrac{m 50^2}{2}=\dfrac{mv^2}{2}$

on solving we get

$v=10\sqrt {33}$ m/s

A gun is mounted on a stationary rail road car. The mass of the car, the gun, the shells and the operator is 50m, where m is the mass of one shell. Two shells are fired one after the other along same horizontal line in same direction. If the muzzle velocity (velocity with respect to gun) of the shells is 200 m/s, then find the speed of the car after second shot.

Centre of mass of velocity of shell $= 0$ (initially at rest)

Therefore , $0 = 49 m \times V + m \times 200$

$V = \dfrac{-200}{49} m/s$

$\dfrac{200}{49} m/s$ towards left .

Another shell is fire , then the velocity of the car , with respect to platform is

$V' = \dfrac{200}{49} m/s$ towards left.

Another shell is fired , then the velocity of the car with respect to platform is

$v' = \dfrac{200}{48} m/s$ towards left.

Now , the velocity of the car with respect to earth is $(\dfrac{200}{49} + \dfrac{200}{48}) = \dfrac{19400}{2352} = 8.24 m/s$ towards left.

A ball is dropped from a height, if it takes $0.2\,s$ to cross the last $6\,m$ before hitting the ground, find the height from which it is dropped. Take $g=10\,m/s^{2}$ (AS1)

Given,

Displacement, $s=6\,m$

Time, $t=0.2\,sec$

Apply kinematic equation of motion

$s=ut+\dfrac{1}{2}a{{t}^{2}}$

$6=u(0.2)+\dfrac{1}{2}\times 10\times {{(0.2)}^{2}}$

$u=29\,m/s$

$v=u+at$

$v=29+10\times 0.2=31\,m/s$

It falls freely from some height $h$ , where initial velocity, $U=0$

Apply second kinematic equation

${{V}^{2}}-{{U}^{2}}=2gh\ \ \ \$

$h=\dfrac{{{V}^{2}}}{2g}=\dfrac{{{31}^{2}}}{2\times 10}=48.05\,m$

From $48.05\,m$ ball is dropped.

b) Two tall building face each other and are at a distance of 180 m from each other. With what velocity must a ball be thrown horizontally from a window 55 m above the ground in one building, so that it enters a window 10.9 m above the ground in the second building?

Let the initial vel of ball be

$v$ and time taken for its light be

$t$

Then,

Also, in vertical displacement

$(10.9-55)=(0)t-\dfrac{g}{2}t^{2}$ [initial vertical component is zero]

$\Rightarrow \sqrt{-\dfrac{44.1\times 2}{-9.8}}=t$

$\Rightarrow 3s=t$

From

$(i)$ and

$(ii)$

$v(3)=180$

$v=60\ m/s$

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From the top of a tower $100\ m$ in flight ball is dropped and the same time another ball is projected vertically upward from the ground with a velocity of $25\ m/sec$ . Find when and where the two balls will meet? $(g=9.8\ m/sec^{2})$

Let take distance covered by the stone thrown upwards to meet the other as $x$ . Then the other stone covers a distance of $(100-x)\ m$ .

Let the ball dropped down be $b_{1}$ . So the ball thrown up is $b_{2}$ .

For ${{b}_{1}}$

$d=100-x$

$g=9.8$

$u=0$

Now, from equation of motion

$s=ut+\frac{1}{2}a{{t}^{2}}$

$100-x=\frac{1}{2}\times 9.8{{t}^{2}}$

$100-x=4.9{{t}^{2}}.....(I)$

For ${{b}_{2}}$

$g=-9.8$

$u=25\,m/s$

Now, from equation of motion

$s=ut+\frac{1}{2}g{{t}^{2}}$

$x=25t-4.9{{t}^{2}}....(II)$

Now, adding equation (I) and (II)

$100-x+x=4.9{{t}^{2}}+25t-4.9{{t}^{2}}$

$100=25t$

$t=4\,s$

Now, putting the value of t in equation (I)

$100-x=4.9\times 16$

$-x=78.4-100$

$x=21.6\,m$

Hence, they meet at $21.6\ m$ from ground after $4\ s$

The velocity time graph of particle moving along a straight line is shown in the figure in the time interval from t= $0$ t= $8$ second answer the following three questions.

we know that

$velocity × time = distance$

in velocity time graph area covered by velocity time gives the distance

hence

area covered $I$ triangle 1st I.e. $t=0 to t= 2sec$

so distance covered

$= 2 ×\dfrac{20 }{2}$

$=20m$

2) $20× 1 +3×20 + 5×10$

$= 20 +60 +50$

$=130m$

A particle is projected up with an initial velocity of $80$ the ball will be at a height of $96$ from the ground after

Initial Velovity = U₀ = 80feet/second.

Height = H = 80 feet.

Acceleration due to gravity = g = 32 feet/s².

Using kinematic 2^nd equation.

H = U₀t + at²/2.

96 = 80t -32t²/2.

t²-5t +6 =0.

Solving above equation We get

t = 2seconds and 3 seconds.

It means when particle goes to upward direction then at 2seconds it is at 96 feet height.

and at the returing time at 3seconds its height is 96feet.

A ball in thrown from a roof top at an angle $45^{\circ}$ above the horizontal. It hits the ground a few second later. At what point during its motion, does the ball have greatest acceleration?

Acceleration is constant throughout the journey, and is always vertically downward equal to g

In the figure shown, find time of flight of the projectile along the inclined plane and range OP.

Given that,

Angle $\theta ={{45}^{0}}$

Velocity $u=20\sqrt{2}\,m/s$

We know that,

The time of flight

$t=\dfrac{2u\sin \theta }{g\cos \theta }$

$t=\dfrac{2\times 20\sqrt{2}}{10}\times 1$

$t=5.7\,s$

Now, the horizontal range is

$R=\dfrac{{{u}^{2}}\sin \theta }{g\cos \theta }$

$R=\dfrac{20\sqrt{2}\times 20\sqrt{2}}{10}$

$R=80m$

Hence, this is the required solution

The velocity time graph of a body is shown in the Fig. The ratio of the average velocity, during the intervals OC and CB is ?

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A motorcycle has to move with a constant speed on an over bridge which is in the form of a circular arc of radius $R$ and has a total length $L$ . Suppose the motorcycle starts from the highest point.
(b) If the motorcycle goes at speed $1/\sqrt{2}$ times the maximum found in part
(a) where will it lose the contact with the road?

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You are to launch a rocket, from just above the ground, with one of the following initial velocity vectors: (1) $\vec { { v }_{ o } } =20\hat { i } +70\hat { j }$ ,(2) $\vec { { v }_{ o } } =-20\hat { i } +70\hat { j }$ , (3) $\vec { { v }_{ o } } =20\hat { i } -70\hat { j }$ , (4) $\vec { { v }_{ o } } =-20\hat { i } -70\hat { j }$ . In your coordinate system, $x$ runs along level ground and $y$ increases upward. (a) Rank the vectors according to the launch speed of the projectile, greatest first. (b) Rank the vectors according to the time of flight of the projectile, greatest first.

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Shown are several curves Explain with reason, which one amongst them can be possible trajectory traced by a projectile ( neglect air friction).

The trajectory of projectile under gravitational force of earth is conic section or parabolic or elliptical or its part whose focus must be the center of the earth.Only (c) option amongst the given curves show the center of the earth as the focus of trajectory.

A particle moves horizontally in uniform circular motion, over a horizontal $xy$ plane. At one instant, it moves through the point at coordinates (4.00 m, 4.00 m) with a velocity of $5.00\hat { i } m/{ s }^{ 2 }$ and an acceleration of $+12.5\hat { j } m/{ s }^{ 2 }$ . What are the (a) x and (b) y coordinates of the center of the circular path?

When traveling in circular motion with constant speed,

the instantaneous acceleration vector necessarily points toward the center.

Thus, the center is “straight up” from the cited point.
(a) Since the center is “straight up” from (4.00 m, 4.00 m),

the x coordinate of the center is

$4.00 m$ .
(b) To find out “how far up” we need to know the radius. Using

$a=\dfrac{v^2}{r}$ we find

$r=\dfrac { { v }^{ 2 } }{ a } =\dfrac { { (5.00m/s) }^{ 2 } }{ { 12.5m/s }^{ 2 } } =2.00m.$

At the time $t = 0$ , apple 1 is dropped from a bridge onto a roadway beneath the bridge; somewhat later, apple 2 is thrown down from the same height. The figure gives the vertical positions $y$ of the apples versus $t$ during the falling until both apples have hit the roadway. The scaling is set by $t_{s} = 2.0\, s$ . With approximately what speed is apple 2 thrown down?

The

$y$ coordinate of Apple 1 obeys:

$y - y_{10} = v_{10}t -\dfrac{1}{2} g t^{2}$ (where

$v_{10} = 0, y = 0$ when

$t = 2.0\, s$ )

$\Rightarrow 0 - y_{10} = -\dfrac{1}{2}(9.8\,m/s^2)(2\,s)^2$

$\Rightarrow y_{01} = 19.6\, m = y_{20}$ (For Apple 2)

Now, from the graph for the coordinate of Apple 2, it is thrown at

$t_0 = 1.0\, s$ with velocity

$v_{20})$ . It hits the ground at

$t = 2.25\,s$ . Then:

$y - y_{20} = v_{20} (t- t_0) - \frac{1}{2} g (t - t_0)^{2}$

$\Rightarrow 0 - 19.6\,m/s = v_{20} (2.25\,s- 1.0\,s) - \frac{1}{2} (9.8\,m/s^2) (2.25\, - 1.0\,s)^{2}$

$\therefore v_{20} = -9.56\,m/s$

Thus, the apple 2 is thrown downward with a velocity

$9.56\,m/s^2$ .

A car travels with speed $30$ km $h^{-1}$ for $30$ minute and then with speed $40$ km $h^{-1}$ for one hour. Find: The average speed of car

$Distance$ =

$Speed \times Time$

$Distance$ =

$30 \times 0.5$ +

$40 \times 1$ =

$55km$

$Time$ =

$1 h$ +

$30min$ =

$1.5h$

$Average\ speed$ =

$\dfrac{Distance}{Time}$ =

$36.67kmh^{-1}$

A point moves along an arc of a circle of radius $R$ . Its velocity depends on the distance covered $s$ $\nu=a \sqrt s$ , where $a$ is a constant. Find the angle $\alpha$ between the vector of the total acceleration and the vector of velocity as a function of $s$

From the equation

$\nu=a \sqrt s$

$w_t=\dfrac{d \nu}{dt}=\dfrac{a}{2 \sqrt s}=\dfrac{a}{2 \sqrt s}a\sqrt s=\dfrac{a^2}{2}$ and

$w_n=\dfrac{\nu^2}{R}=\dfrac{a^2 s}{R}$

$w_t$ is a positive constant, the speed of the particle increases with time, and the tangential acceleration vector and velocity coincides in direction

Hence the angle between

$\vec \nu$ and

$\vec w$ is equal to between

$w_t \hat u_1$ an

$\vec w$ and

$\alpha$ can be found by mean of the formula :

$\tan \alpha =\dfrac{|w_n|}{|w_t|}=\dfrac{a^2 s/R}{a^2/2}=\dfrac{2s}{R}$

A person standing at the top of a hemispherical rock of radius $R$ kicks a ball (initially at rest on the top of the rock) to give it horizontal velocity $\vec{V}_{i}$ as shown in above figure. (a) What must be its minimum initial speed if the ball is never to hit the rock after it is kicked? (b) With this initial speed, how far from the base of the rock does the ball hit the ground?

Measure heights above the level ground. The elevation

$y_{b}$ of the ball follows:

$y_{b}=R+0-\frac{1}{2}gt^{2}$
with

$x=v_{i}t$ so

$y_{b}=R-\frac{gx^{2}}{2v_{i}^{2}}$
(a) The elevation

$y_{r}$ of points on the rock is described by

$y_{r}^{2}+x^{2}=R^{2}$
We will have

$y_{b}=y$ , at

$x=0$ , but for all other

$x$ we require the ball to be above the rock's surface as in

$y_{b}>y_{r}$ . Then

$y_{b}^{2}+x^{2}>R^{2}$ :

$\left ( R-\frac{gx^{2}}{2v_{i}^{2}} \right )+x^{2}>R^{2}$

$R^{2}-\frac{gx^{2}R}{v_{i}^{2}}+\frac{g^{2}x^{4}}{4v_{i}^{2}}+x^{2}>R^{2}$

$\frac{g^{2}x^{4}}{4v_{i}^{4}}+x^{2}>\frac{gx^{2}R}{v_{i}^{2}}$
If this inequality is satisfied for

$x$ approaching zero, it will be true for all

$x$ . If the ball’s parabolic trajectory has large enough radius of curvature at the start, the ball will clear the whole rock:

$1>\frac{gR}{v_{i}^{2}}$ , so

$v_{i}>\sqrt{gR}$
(b) With

$v_{i}=\sqrt{gR}$ and

$y_{b}=0$ , we have

$0=R-\frac{gx^{2}}{2gR}$
or

$x=R\sqrt{2}$ . The diatnce from the rock's base is

$x-R=(\sqrt{2}-1)R$

A novice golfer on the green takes three strokes to sink the ball. The successive displacements of the ball are $4.00 m$ to the north, $2.00 m$ northeast, and $1.00 m$ at $30.0^{0}$ west of south (Fig. below). Starting at the same initial point, an expert golfer could make the hole in what single displacement?

We first tabulate the three strokes of the novice golfer, with the

$x$ direction corresponding to East and the

$y$ direction corresponding to North. The sum of the displacement in each of the directions is shown below.
The “hole-in-one” single displacement is then

$|\vec{R}|=\sqrt{|x|^{2}+|y|^{2}}=\sqrt{(0.914m)^{2}+(4.55m)^{2}}=4.64m$
The angle of the displacement with the horizontal is

$\theta=\tan^{-1}\left ( \dfrac{y}{x} \right )=\tan^{-1}\left ( \dfrac{4.55m}{0.914m} \right )=78.6^{0}$
The expert golfer would accomplish the hole in one with the displacement

$4.64 m$ at

$78.6^{0} N$ of

$E$ .

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A ball rolls off the edge of a horizontal table top 4 m high. If it strikes the floor at a point 5 m horizontally away from the edge of the table, what was its speed at the instant it left the table?

Using $\displaystyle h= \frac{1}{2}gt^{2},$ we have

$\displaystyle h_{AB}= \frac{1}{2}gt_{AC}^{2}$

or $\displaystyle t_{AC}= \sqrt{\frac{2h_{AB}}{g}}$

$\displaystyle = \sqrt{\frac{2\times 4}{9.8}}= 0.9s$

Further, $\displaystyle BC= vt_{AC}.$

or $\displaystyle v= \frac{BC}{t_{AC}}= \frac{5.0}{0.9}= 5.55 m/s$

The trajectories of the motion of 3 particles are shown in the figure. Match the entries of column I with the entries of column II. Neglect air resistance.

Since time of flight depends on

$u_v$ and vertical height is same for all therefore,

$u_v$ will be same for all. Hence, time of flight will be same.
A->4
B->4
(C) Since horizontal range is greatest for C therefore,

$u_h$ will be greatest for (C)
C->3
(D)Since

$u_v$ is same for all but

$u_h$ is least for (A) therefore, launch speed will be least for (A)
D->1

A projectile is fired horizontally with a velocity of $98 m/s$ from the top of a hill $490 m$ high. Find the horizontal range of the projectile.

Given :

$u_x = 98 m/s$

$a_y = g = 9.8 m/s^2$

Height of the hill

$S_y = h = 490$ m

$\therefore$ Time of flight

$T = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 (490)}{9.8}} = 10$ s

$\therefore$ Horizontal range

$R = u_x T = 98 \times 10 = 980m$

A ball is thrown vertically upwards. After some time, it returns to the thrower. Draw the velocity-time graph and speed-time graph.

Initially ball will go up with decreasing speed.

At highest point velocity will be zero.

Then ball will fall down ward with increasing speed.

Taking upward direction as (+)ve.

Hence, solve.

A person standing on the top of a cliff $171\ ft$ high has to throw a packet to his friend standing on the ground $228\ ft$ horizontally away. If he throws the packet directly aiming at the friend with a speed of $15\cdot 0\ ft/s$ , how short will the packet fall?

The straight line distance from A to B is

$\sqrt{171^2 + 228^2}$ = 285

feet.

Packet is thrown along the direction AB with speed 15 ft/s.

If we resolve this speed into vertical and horizontal components,

we get vertical component as 9 ft/s

( 15 sinθ =

$\frac{15 \times 171}{285}$ = 9 ft/s)

and horizontal component as 12 ft/s

$( 15 cosθ = \frac{15 \times 228}{285}$ = 12 ft/s )

The time to reach the ground for the packet is decided by gravity.

Hence we will find the time taken to reach the ground using vertical component.

we need to use the equation of motion, " S =

$ut + (\frac{1}{2})gt^2$ "

171 =

$9 \times t + (\frac{1}{2}) \times 9.8 \times 3.281 \times t^2$

$= 9 \times t + 16.08 \times t^2$ ...................(1)

(acceleration due to gravity is

$ft/s^2$ . 1 m = 3.281 ft)

solving the quadratic equation (1) for t, we get t = 3 s.

Hence horizontal distance travelled by the packet =

$12 \times$ 3 = 36 ft.

Packet will fall at a distance 36 ft from O,

short of 192 ft to B ( 228-36 = 192 ft )

1108351_1031195_ans_a6e1a615dc104b0c8bbb788b0506476e.jpg

A ball A is a projected from the ground such that its horizontal range is maximum. Another ball left it is dropped from a height equal from a height equal to the maximum range of A?

In first ball,

$T=\dfrac { 2usin\theta }{ g }$

$T\left( A \right) =\dfrac { 2usin{ 45 }^{ o } }{ g }$

$T\left( A \right) =\dfrac { 1.41u }{ g }$

${ R }_{ max }=\dfrac { { u }^{ 2 } }{ g }$

In second ball,

$s=ut+\dfrac { 1 }{ 2 } { gt }^{ 2 }$

${ R }_{ max }=ut+\dfrac { 1 }{ 2 } g{ T_{B}}^{ 2 }$

$T_B=\sqrt { \dfrac { { 2R }_{ max } }{ g } } =\sqrt { \dfrac { 2{ u }^{ 2 } }{ { g }^{ 2 } } } =\sqrt { \dfrac { 2{ u } }{ { g } } }$

$\dfrac { T\left( A \right) }{ T\left( B \right) } =\dfrac { 1.41 }{ \sqrt { 2 } } =1$

An airplane flying horizontally at a constant speed of $350 \,km/h$ over level ground releases a bundle of food supplies. Ignore the effect of the air on the bundle. What are the bundles initial (a) vertical and (b) horizontal components of velocity? (c) What is its horizontal component of velocity just before hitting the ground? (d) If the airplanes speed were, instead, $450 \,km/h$ , would the time of fall be longer, shorter, or the same?

From the given data,

$v_{oy}= 0$ km/h,

$v_{ox} = 350$ km/h,

(c) What is its horizontal component of velocity just before hitting the ground? There is no significant horizontal acceleration so

$v_x = 350$ km/h,

(d) If the airplane's speed were, instead, 450 km/h, would the time of fall be longer, shorter, or the same? Same. Time of fall is determined by vertical acceleration and vertical initial velocity

A disk, initially rotating at 120 rad/s , is slowed down with a constant angular acceleration of magnitude 4.0 rad/ $s^{2}$ . (a) How much time does the disk take to stop (b) Through what angle does the disk rotate during that time

1967797_1765994_ans_2d217231bd88426097aec17e60625590.jpg

A certain airplane has a speed of $290.0 km/h$ and is diving at an angle of $30.0$ below the horizontal and when the pilot releases a radar decoy (Fig. 4-33). The horizontal distance between the release point and the point where the decoy strikes the ground is $d =700 m.$ (a) How long is the decoy in the air? (b) How high was the release point?

We adopt the positive direction choices used in the textbook so that equations such as
Eq. are directly applicable. The coordinate origin is at ground level directly below
the release point. We write

${ theta }_{ 0 } = –30.0°$ since the angle shown in the figure is measured
clockwise from horizontal. We note that the initial speed of the decoy is the plane’s speed
at the moment of release:

${ v }_{ 0 } = 290 km/h$ , which we convert to SI units: $$(290)(1000/3600)
= 80.6 m/s.$$

(a) We use Eq. to solve for the time:

$\Delta x=\left( { v }_{ 0 }\cos { { \theta }_{ 0 } } \right) t\Rightarrow t=\frac { 700m }{ (80.6m/s)\cos { \left( { -30.0 }^{ 0 } \right) } } =10.0s.$
(b) And we use Eq. 4-22 to solve for the initial height

${ y }_{ 0 }$ :

$y-{ y }_{ 0 }=\left( { v }_{ 0 }\sin { { \theta }_{ 0 } } \right) t-\frac { 1 }{ 2 } g{ t }^{ 2 }\Rightarrow 0-{ y }_{ 0 }=(-40.3m/s)(10.0s)-\frac { 1 }{ 2 } (9.80/{ s }^{ 2 })(10.0{ s }^{ 2 })$
which yields

${ y }_{ 0 } = 897 m.$

A rifle that shoots bullets at $460 m/s$ is to be aimed at a target $45.7 m$ away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at the end of the rifle (the initial point for the bullet as it begins projectile motion in the sense of § 4-5), and we let

${ \theta }_{ 0 }$ be the firing angle. If the target is a distance

$d$ away, then its coordinates are

$x = d$ ,

$y = 0$ .
The projectile motion equations lead to

$d=({ v }_{ 0 }{ cos\theta }_{ 0 })t$

$0={ v }_{ 0 }t{ sin\theta }_{ 0 }-\frac { 1 }{ 2 } g{ t }^{ 2 }$
The setup of the problem is shown in the figure.
The time at which the bullet strikes the target is

$t=d/({ v }_{ 0 }{ cos\theta }_{ 0 })$ . Eliminating

$t$ leads to

$2{ v }_{ 0 }^{ 2 }{ sin\theta }_{ 0 }{ cos\theta }_{ 0 }-gd=0.$
Using

${ sin\theta }_{ 0 }{ cos\theta }_{ 0 }=\frac { 1 }{ 2 } sin(2{ \theta }_{ 0 })$ , we obtain

${ v }_{ 0 }^{ 2 }{ sin(2\theta }_{ 0 })=gd\Rightarrow sin(2{ \theta }_{ 0 })=\frac { gd }{ { v }_{ 0 }^{ 2 } } =\frac { (9.80m/{ s }^{ 2 })(45.7m) }{ { (460m/s) }^{ 2 } }$
which yields

$sin({ 2\theta }_{ 0 })=2.11\times { 10 }^{ -3 }$ , or

${ \theta }_{ 0 }=0.0606^{ o }$ . If the gun is aimed at a point a
distance

$l$ above the target, then

$tan{ \theta }_{ 0 }=l/d$ so that

$l=d{ tan\theta }_{ 0 }=(45.7m)tan({ 0.0606 }^{ o })=0.0484m=4.84cm.$
Note that due to the downward gravitational acceleration, in order for then bullet to strike the target, the gun must be aimed at a point slightly above the target.
1643275_1770634_ans_4d7aedfcbd7f49b0b5bc3768e2cd968f.png

1643275_1770634_ans_4d7aedfcbd7f49b0b5bc3768e2cd968f.png

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wall to break apart the wall. The machine was not placed near the wall because then arrows could reach it from the castle wall. Instead, it was positioned so that the stone hit the wall during the second half of its flight. Suppose a stone is launched with a speed of ${ v }_{ 0 }=28.0 m/s$ and at an angle of ${ \theta }_{ 0 }=40.0$ . What is the speed of the stone if it hits the wall (a) just as it reaches the top of its parabolic path and (b) when it has descended to half that height? (c) As a percentage, how much faster is it moving in part (b) than in part (a)?

(a) Since the y-component of the velocity of the stone at the top of its path is zero, its
speed is

$v=\sqrt { { v }_{ x }^{ 2 }+v_{ y }^{ 2 } } ={ v }_{ x }={ v }_{ 0 }cos{ \theta }_{ 0 }=(28.0m/s)cos{ 40.0 }^{ o }=21.4m/s.$
(b) Using the fact that

${ v }_{ y }=0$ at the maximum height

${ y }_{ max }$ , the amount of time it takes for
the stone to reach

${ y }_{ max }$ is given by Eq. 4-23:

$0={ v }_{ y }={ v }_{ 0 }{ sin\theta }_{ 0 }-gt\Rightarrow t=\frac { { v }_{ 0 }{ sin\theta }_{ 0 } }{ g } .$
Substituting the above expression into Eq. 4-22, we find the maximum height to be

${ y }_{ max }=\left( { v }_{ 0 }{ sin\theta }_{ 0 } \right) t-\frac { 1 }{ 2 } g{ t }^{ 2 }={ v }_{ 0 }{ sin\theta }_{ 0 }\left( \frac { { v }_{ 0 }{ sin\theta }_{ 0 } }{ g } \right) -\frac { 1 }{ 2 } g{ \left( \frac { { v }_{ 0 }{ sin\theta }_{ 0 } }{ g } \right) }^{ 2 }\frac { { v }_{ 0 }^{ 2 }{ sin }^{ 2 }{ \theta }_{ 0 } }{ 2g } .$
To find the time the stone descends to max

$y={ y }_{ max }/2$ , we solve the quadratic equation given
in Eq. 4-22:

$y=\frac { 1 }{ 2 } { y }_{ max }=\frac { { v }_{ 0 }^{ 2 }{ sin }^{ 2 }{ \theta }_{ 0 } }{ 4g } =({ v }_{ 0 }{ sin\theta }_{ 0 })t-\frac { 1 }{ 2 } g{ t }^{ 2 }\Rightarrow { t }_{ +- }=\frac { ({ 2 }_{ - }^{ + }\sqrt { 2 } ){ v }_{ 0 }{ sin\theta }_{ 0 } }{ 2g } .$
Choosing

$t={ t }_{ + }$ (for descending), we have

${ v }_{ x }={ v }_{ 0 }{ cos\theta }_{ 0 }=(28.0m/s){ cos40.0 }^{ o }=21.4m/s$

${ v }_{ y }={ v }_{ 0 }{ sin\theta }_{ 0 }-g\frac { (2+\sqrt { 2 } ){ v }_{ 0 }{ sin\theta }_{ 0 } }{ 2g } =-\frac { \sqrt { 2 } }{ 2 } { v }_{ 0 }{ sin\theta }_{ 0 }=-\frac { \sqrt { 2 } }{ 2 } (28.0m/s)sin{ 40.0 }^{ o }=-12.7m$
Thus, the speed of the stone when

$y={ y }_{ max }/2$ is

$v=\sqrt { { v }_{ x }^{ 2 }+{ v }_{ y }^{ 2 } } =\sqrt { { (21.4m/s) }^{ 2 }+{ (-12.7m/s) }^{ 2 } } =24.9m/s.$
(c) The percentage difference is

$\frac { 24.9m/s-21.4m/s }{ 21.4m/s } =0.163=16.3%.$

Entering his dorm room, a student tosses his book bag to the right and upward at an angle of $45^{0}$ with the horizontal (Fig. above). Air resistance does not affect the bag. The bag moves through point $A$ immediately after it leaves the students hand, through point $B$ at the top of its flight, and through point $C$ immediately before it lands on the top bunk bed.
(i) Rank the following horizontal and vertical velocity components from the largest to the smallest.
(a) $v_{Ax}$ (b) $v_{Ay}$ (c) $v_{Bx}$ (d) $v_{By}$ (e) $v_{Cy}$ .
Note that zero is larger than a negative number. If two quantities are equal, show them as equal in your list. If any quantity is equal to zero, show that fact in your list.
(ii) Similarly, rank the following acceleration components.
(a) $a_{Ax}$ (b) $a_{Ay}$ (c) $a_{Bx}$ (d) $a_{By}$ (e) $a_{Cy}$ .

(i) The

$45^{0}$ angle means that at point

$A$ the horizontal and vertical velocity components are equal. The horizontal velocity component is the same at

$A, B$ , and

$C$ . The vertical velocity component is zero at

$B$ and negative at

$C$ . The assembled answer is

$a = b = c > d = 0 > e$ .
(ii) The

$x$ component of acceleration is everywhere zero and the y component is everywhere

$−9.80 m/s^{2}$ . Then we have

$a = c = 0 > b = d = e$ .

Three schoolboys, Sam, Jhon, and Nick, are on merry-go-round. Sam and John occupy diametrically opposite points on a merry-go-round of radius $R$ . The position of the boys at the initial instant are shown in fig
Considering that the merry-go-round touch each other and rotate in the same direction at the same angular velocity $\omega$ determine the nature of motion of Nick from John's point of view and of Sam from Nick's point of view.

Let the merry-go-round turn through a certain angle

$\varphi$ Fig. We construct a point

$O (OA=R)$ such that points

$O, S,A$ and

$J$ lie on the same straight line. Then it is clear that

$ON=R+ r$ at any instant of time. Besides Point

$O$ is at rest relative to John (Sam is always opposite to John). Therefore, from John's point of view, NIck translates in a circle of radius

$R+r$ with thecentere at point

$O$ which moves relative to the ground in a circle of radius

$R$ with the centre at point

$A$ . From Nick's point of view, sam translates in a which is at rest relative to Nick. However point

$O$ moves relative to the ground in a circle of radius

$r$ with the centre at point

$B$ .

1808289_1860277_ans_1c8a001f3a3947709037fa71bc1fceda.png

A $1.25-kg$ wooden block rests on a table over a large hole as in Figure $P9.71$ . A $5.00-g$ bullet with an initial velocity $v_i$ is fired upward into the bottom of the block and remains in the block after the collision. The block and bullet rise to a maximum height of $22.0\ cm$ .
Describe how you would find the initial velocity of the bullet using ideas you have learned in this chapter.

Given,

Mass of wooden block is

$1.25kg$

mass of bullet is

$5g=0.0$

maximum height is

$22cm=0.22m$

We know that final velocity is

${ v }_{ b }=\sqrt { 2gh }$

Substitute all value in above equation

${ v }_{ b }=\sqrt { 2\times 9.8\times 0.22 } \\ { v }_{ b }=2.08m/s$

as this collision is inelastic collision therefore momentum will be conserved before and after collision.

${ m }_{ 1 }{ v }_{ i }+{ m }_{ 2 }{ v }_{ 2 }=({ m }_{ 1 }+{ m }_{ 2 }){ v }_{ b }$

Substitute all value in above equation

$0.005\times { v }_{ i }+1.25\times 0=(0.005+1.25)2.08\\ { v }_{ i }=\frac { 2.61 }{ 0.005 } \\ { v }_{ i }=522.08m/s$

Hence,

Initial velocity of the bullet is

$522.08m/s$

Motion In A Plane - Class 11 Medical Physics - Extra Questions

A ball is thrown from the top of a tower with an initial velocity of 10 $ms^{-1}$ at an angle of $30^0$ above the horizontal. It hits the ground at a distance of 17.3 m from the base of the tower. Calculate the height of the tower. (g = 10 $ms^{-2}$ )

A ball is thrown upwards from the top of a tower 40 m high with a velocity of 10 m/s. Find the time when it strikes the ground (Take g = 10 $m/s^{2}$ ).

The resultant of two like parallel forces 12 N and 7 N is ________.

$\displaystyle \bigtriangleup ABC$ is right-angled at C If AC = 5 cm and BC=12 cm find the length of AB

A car is moving with speed $30\ m/sec$ on a circular path of radius $500\ m$ . Its speed is increasing at the rate of $2\ m/sec^{2}$ . What is the net acceleration of the car at that moment (i.e the vector sum of tangential and radial acceleration)?

Two forces $5$ kg wt and $10$ kg wt are acting with an inclination of $120^o$ between them. what is the angle the resultant makes with $10$ kg wt?

Three point charges $+Q, + 2Q$ and $-3Q$ are placed at the vertices of an equilateral triangle ABC of side L.if these charges are displaced to the mid-points $A_1, B_1$ and $C_1$ respectively find the amount of the work done in shifting the charges to the new locations.

Find a unit vector parallel to the resultant of the vectors $\bar{A}=2\hat{i}+3\hat{J}+4\hat{k}$ and $\bar{B}=3\hat{i}-5\hat{J}$ .

A ball is thrown from the top of a building 45 m high with a speed 20 m s $^{-1}$ above the horizontal at an angle of $30^o$ . Find
a. The time taken by the ball to reach the ground.
b. The speed of ball just before it touches the ground.

Two bullets are fired with horizontal velocities of $50m/s$ and $100m/s$ from two guns at a height of $19.6m$ (a) Will both the bullets strike the ground? (b) If yes, then after how much time and which bullet will strike first? (c) What would be the path of the bullets? ( $g=9.8m/{s}^{2}$ )

Two forces each of $10\ N$ act at an angle ${60}^{o}$ with each other. The magnitude & direction of the resultant with respect to one of the vectors is

A projectile fired from the top of a $40\ m$ high cliff with an initial speed of $50\ m/s$ at an unknown angle. Find its speed when it hits the ground. $(g=10\ {m/s}^{2})$

A ball is dropped from a height, if it takes $0.2\,s$ to cross the last $6\,m$ before hitting the ground, find the height from which it is dropped. Take $g=10\,m/s^{2}$ (AS1)

b) Two tall building face each other and are at a distance of 180 m from each other. With what velocity must a ball be thrown horizontally from a window 55 m above the ground in one building, so that it enters a window 10.9 m above the ground in the second building?

From the top of a tower $100\ m$ in flight ball is dropped and the same time another ball is projected vertically upward from the ground with a velocity of $25\ m/sec$ . Find when and where the two balls will meet? $(g=9.8\ m/sec^{2})$

The velocity time graph of particle moving along a straight line is shown in the figure in the time interval from t= $0$ t= $8$ second answer the following three questions.

A particle is projected up with an initial velocity of $80$ the ball will be at a height of $96$ from the ground after

A ball in thrown from a roof top at an angle $45^{\circ}$ above the horizontal. It hits the ground a few second later. At what point during its motion, does the ball have greatest acceleration?

In the figure shown, find time of flight of the projectile along the inclined plane and range OP.

The velocity time graph of a body is shown in the Fig. The ratio of the average velocity, during the intervals OC and CB is ?

Shown are several curves Explain with reason, which one amongst them can be possible trajectory traced by a projectile ( neglect air friction).

A car travels with speed $30$ km $h^{-1}$ for $30$ minute and then with speed $40$ km $h^{-1}$ for one hour. Find: The average speed of car

A point moves along an arc of a circle of radius $R$ . Its velocity depends on the distance covered $s$ $\nu=a \sqrt s$ , where $a$ is a constant. Find the angle $\alpha$ between the vector of the total acceleration and the vector of velocity as a function of $s$

A ball rolls off the edge of a horizontal table top 4 m high. If it strikes the floor at a point 5 m horizontally away from the edge of the table, what was its speed at the instant it left the table?

The trajectories of the motion of 3 particles are shown in the figure. Match the entries of column I with the entries of column II. Neglect air resistance.

A projectile is fired horizontally with a velocity of $98 m/s$ from the top of a hill $490 m$ high. Find the horizontal range of the projectile.

A ball is thrown vertically upwards. After some time, it returns to the thrower. Draw the velocity-time graph and speed-time graph.

A person standing on the top of a cliff $171\ ft$ high has to throw a packet to his friend standing on the ground $228\ ft$ horizontally away. If he throws the packet directly aiming at the friend with a speed of $15\cdot 0\ ft/s$ , how short will the packet fall?

A ball A is a projected from the ground such that its horizontal range is maximum. Another ball left it is dropped from a height equal from a height equal to the maximum range of A?

A disk, initially rotating at 120 rad/s , is slowed down with a constant angular acceleration of magnitude 4.0 rad/ $s^{2}$ . (a) How much time does the disk take to stop (b) Through what angle does the disk rotate during that time

A rifle that shoots bullets at $460 m/s$ is to be aimed at a target $45.7 m$ away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

Class 11 Medical Physics Extra Questions

Motion In A Plane - Class 11 Medical Physics - Extra Questions

A ball is thrown from the top of a tower with an initial velocity of 10 ms−1ms^{-1} at an angle of 30030^0 above the horizontal. It hits the ground at a distance of 17.3 m from the base of the tower. Calculate the height of the tower. (g = 10 ms−2ms^{-2})

A ball is thrown upwards from the top of a tower 40 m high with a velocity of 10 m/s. Find the time when it strikes the ground (Take g = 10 m/s2m/s^{2}).

The resultant of two like parallel forces 12 N and 7 N is ________.

△ABC \displaystyle \bigtriangleup ABC is right-angled at C If AC = 5 cm and BC=12 cm find the length of AB

A car is moving with speed 30 m/sec30\ m/sec on a circular path of radius 500 m500\ m. Its speed is increasing at the rate of 2 m/sec22\ m/sec^{2}. What is the net acceleration of the car at that moment (i.e the vector sum of tangential and radial acceleration)?

Two forces 55kg wt and 1010kg wt are acting with an inclination of 120^o120^o between them. what is the angle the resultant makes with 1010kg wt?

Three point charges +Q, + 2Q+Q, + 2Q and -3Q-3Q are placed at the vertices of an equilateral triangle ABC of side L.if these charges are displaced to the mid-points A_1, B_1A_1, B_1 and C_1C_1 respectively find the amount of the work done in shifting the charges to the new locations.

Find a unit vector parallel to the resultant of the vectors \bar{A}=2\hat{i}+3\hat{J}+4\hat{k}\bar{A}=2\hat{i}+3\hat{J}+4\hat{k} and \bar{B}=3\hat{i}-5\hat{J}\bar{B}=3\hat{i}-5\hat{J}.

A ball is thrown from the top of a building 45 m high with a speed 20 m s^{-1}^{-1} above the horizontal at an angle of 30^o30^o. Finda. The time taken by the ball to reach the ground.b. The speed of ball just before it touches the ground.

Two forces each of 10\ N10\ N act at an angle {60}^{o}{60}^{o} with each other. The magnitude & direction of the resultant with respect to one of the vectors is

A projectile fired from the top of a 40\ m40\ m high cliff with an initial speed of 50\ m/s50\ m/s at an unknown angle. Find its speed when it hits the ground. (g=10\ {m/s}^{2})(g=10\ {m/s}^{2})

A ball is dropped from a height, if it takes 0.2\,s0.2\,s to cross the last 6\,m6\,m before hitting the ground, find the height from which it is dropped. Take g=10\,m/s^{2}g=10\,m/s^{2} (AS1)

b) Two tall building face each other and are at a distance of 180 m from each other. With what velocity must a ball be thrown horizontally from a window 55 m above the ground in one building, so that it enters a window 10.9 m above the ground in the second building?

From the top of a tower 100\ m100\ m in flight ball is dropped and the same time another ball is projected vertically upward from the ground with a velocity of 25\ m/sec25\ m/sec. Find when and where the two balls will meet? (g=9.8\ m/sec^{2})(g=9.8\ m/sec^{2})

The velocity time graph of particle moving along a straight line is shown in the figure in the time interval from t=00 t=88second answer the following three questions.

A particle is projected up with an initial velocity of 8080 the ball will be at a height of 9696 from the ground after

A ball in thrown from a roof top at an angle 45^{\circ}45^{\circ} above the horizontal. It hits the ground a few second later. At what point during its motion, does the ball have greatest acceleration?

In the figure shown, find time of flight of the projectile along the inclined plane and range OP.

The velocity time graph of a body is shown in the Fig. The ratio of the average velocity, during the intervals OC and CB is ?

Shown are several curves Explain with reason, which one amongst them can be possible trajectory traced by a projectile ( neglect air friction).

A car travels with speed 3030 km h^{-1}h^{-1} for 3030 minute and then with speed 4040 km h^{-1}h^{-1} for one hour. Find: The average speed of car

A point moves along an arc of a circle of radius RR. Its velocity depends on the distance covered ss \nu=a \sqrt s\nu=a \sqrt s, where aa is a constant. Find the angle \alpha\alpha between the vector of the total acceleration and the vector of velocity as a function of ss

A ball rolls off the edge of a horizontal table top 4 m high. If it strikes the floor at a point 5 m horizontally away from the edge of the table, what was its speed at the instant it left the table?

The trajectories of the motion of 3 particles are shown in the figure. Match the entries of column I with the entries of column II. Neglect air resistance.

A projectile is fired horizontally with a velocity of 98 m/s98 m/s from the top of a hill 490 m490 m high. Find the horizontal range of the projectile.

A ball is thrown vertically upwards. After some time, it returns to the thrower. Draw the velocity-time graph and speed-time graph.

A person standing on the top of a cliff 171\ ft171\ ft high has to throw a packet to his friend standing on the ground 228\ ft228\ ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15\cdot 0\ ft/s15\cdot 0\ ft/s, how short will the packet fall?

A ball A is a projected from the ground such that its horizontal range is maximum. Another ball left it is dropped from a height equal from a height equal to the maximum range of A?

A disk, initially rotating at 120 rad/s , is slowed down with a constant angular acceleration of magnitude 4.0 rad/ s^{2} s^{2} . (a) How much time does the disk take to stop (b) Through what angle does the disk rotate during that time

A rifle that shoots bullets at 460 m/s460 m/s is to be aimed at a target 45.7 m45.7 m away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

Class 11 Medical Physics Extra Questions

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A ball is thrown from the top of a tower with an initial velocity of 10 $ms^{-1}$ at an angle of $30^0$ above the horizontal. It hits the ground at a distance of 17.3 m from the base of the tower. Calculate the height of the tower. (g = 10 $ms^{-2}$ )

A ball is thrown upwards from the top of a tower 40 m high with a velocity of 10 m/s. Find the time when it strikes the ground (Take g = 10 $m/s^{2}$ ).

$\displaystyle \bigtriangleup ABC$ is right-angled at C If AC = 5 cm and BC=12 cm find the length of AB

A car is moving with speed $30\ m/sec$ on a circular path of radius $500\ m$ . Its speed is increasing at the rate of $2\ m/sec^{2}$ . What is the net acceleration of the car at that moment (i.e the vector sum of tangential and radial acceleration)?

Two forces $5$ kg wt and $10$ kg wt are acting with an inclination of $120^o$ between them. what is the angle the resultant makes with $10$ kg wt?

Three point charges $+Q, + 2Q$ and $-3Q$ are placed at the vertices of an equilateral triangle ABC of side L.if these charges are displaced to the mid-points $A_1, B_1$ and $C_1$ respectively find the amount of the work done in shifting the charges to the new locations.

Find a unit vector parallel to the resultant of the vectors $\bar{A}=2\hat{i}+3\hat{J}+4\hat{k}$ and $\bar{B}=3\hat{i}-5\hat{J}$ .

A ball is thrown from the top of a building 45 m high with a speed 20 m s $^{-1}$ above the horizontal at an angle of $30^o$ . Find
a. The time taken by the ball to reach the ground.
b. The speed of ball just before it touches the ground.

Two forces each of $10\ N$ act at an angle ${60}^{o}$ with each other. The magnitude & direction of the resultant with respect to one of the vectors is

A projectile fired from the top of a $40\ m$ high cliff with an initial speed of $50\ m/s$ at an unknown angle. Find its speed when it hits the ground. $(g=10\ {m/s}^{2})$

A ball is dropped from a height, if it takes $0.2\,s$ to cross the last $6\,m$ before hitting the ground, find the height from which it is dropped. Take $g=10\,m/s^{2}$ (AS1)

From the top of a tower $100\ m$ in flight ball is dropped and the same time another ball is projected vertically upward from the ground with a velocity of $25\ m/sec$ . Find when and where the two balls will meet? $(g=9.8\ m/sec^{2})$

The velocity time graph of particle moving along a straight line is shown in the figure in the time interval from t= $0$ t= $8$ second answer the following three questions.

A particle is projected up with an initial velocity of $80$ the ball will be at a height of $96$ from the ground after

A ball in thrown from a roof top at an angle $45^{\circ}$ above the horizontal. It hits the ground a few second later. At what point during its motion, does the ball have greatest acceleration?

A car travels with speed $30$ km $h^{-1}$ for $30$ minute and then with speed $40$ km $h^{-1}$ for one hour. Find: The average speed of car

A point moves along an arc of a circle of radius $R$ . Its velocity depends on the distance covered $s$ $\nu=a \sqrt s$ , where $a$ is a constant. Find the angle $\alpha$ between the vector of the total acceleration and the vector of velocity as a function of $s$

A projectile is fired horizontally with a velocity of $98 m/s$ from the top of a hill $490 m$ high. Find the horizontal range of the projectile.

A person standing on the top of a cliff $171\ ft$ high has to throw a packet to his friend standing on the ground $228\ ft$ horizontally away. If he throws the packet directly aiming at the friend with a speed of $15\cdot 0\ ft/s$ , how short will the packet fall?

A disk, initially rotating at 120 rad/s , is slowed down with a constant angular acceleration of magnitude 4.0 rad/ $s^{2}$ . (a) How much time does the disk take to stop (b) Through what angle does the disk rotate during that time

A rifle that shoots bullets at $460 m/s$ is to be aimed at a target $45.7 m$ away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?