non zero acceleration without having varying speed

varying speed without having varying velocity

varying velocity without having varying speed

non zero acceleration without having varying velocity

MCQ Questions for Class 9 Physics Motion Quiz 4

A ball is thrown vertically upwards with a speed of

$10\ ms^{-1}$ from the ground at the bottom of a tower

$200$

$m$ high. Another is dropped vertically downward simultaneously, from the top of a tower. If

$g=10\ ms^{-2}$ the time interval after which the projected body will be at the same level as the dropped body is:

0%
$20\ s$
0%
$25\ s$
0%
$2\sqrt{10}\;s$
0%
$5\ s$

A stone is dropped from the top of a tower of height

$49$

$m$ . Another stone is thrown up vertically with velocity of

$24.5\ m/s$ from the foot of the tower at the same instant. They will meet in a time of

0%
1 s
0%
2 s
0%
0.5 s
0%
0.25 s

Two balls A and B are thrown simultaneously. A vertically upwards at a speed of

$15$ m/s from the ground and B, vertically downwards from a height of

$30$ m at the same speed along the same line of motion. They meet after a time of:

0%
1 s
0%
2 s
0%
3 s
0%
4 s

A body is projected vertically upwards with a velocity

$u$ . It crosses a point in its journey at a height

$h$ meter twice, just after

$1$ and

$7$ seconds .The value of

$u$ in

$ms^{-1}$ is

$(g= 10ms^{-2})$

0%
$50$
0%
$40$
0%
$30$
0%
$20$

Explanation

$h = ut - \dfrac {1}{2}gt^2$

$h = u(1) - \dfrac {1}{2}g = u - \dfrac {g}{2} \ \ \ ..... (1)$

$h = u(7) - \dfrac {1}{2}g (7)^2 = 7 u - \dfrac {49g}{2} \ \ \ ..... (2)$
Equating (1) & (2)

$u - \dfrac {g}{2} = 7u - \dfrac {49g}{2}$

$24g = 6u$

$u = 4g$

$\Rightarrow u = 40\ m/s$

An object may have

0%
varying speed without having varying velocity
0%
varying velocity without having varying speed
0%
non zero acceleration without having varying velocity
0%
non zero acceleration without having varying speed

The displacement (

$x$ ) versus time (

$t$ ) curve for two particles is as shown in figure. Which of the following statements is correct for time interval between

$0$ to

$10$ s?

0%
The particle $A$ is speeding up while $B$ is slowing down.
0%
Both the particles are initially speeding up and then slowing down.
0%
Both the particles are initially slowing down and then speeding up.
0%
Particle $A$ is speeding up first and then slowing down while $B$ is slowing down first and then speeding up.

A stone is thrown vertically up from the ground. It reaches a maximum height of 50 m in 10 s. After what time will it reach the ground from the maximum height?

0%
5 s
0%
10 s
0%
20 s
0%
25 s

Explanation

$t_a = \displaystyle \dfrac{u}{g} = 10 s$

same amount of time it will take to fall down

A body moving with uniform acceleration in a straight line is at points A, B, C, D after successive equal intervals of time. The distance AD is equal to:

0%
BC
0%
2(BC)
0%
3(BC)
0%
4(BC)

Explanation

Let the starting point is O. And the body reach,

A at time t,

B at time 2t.

C at time 3t,

D at time 4t.

Using

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

O to A is

$ut+\dfrac{t^2}{2}$

O to B is

$ut+\dfrac{4t^2}{2}$

O to C is

$ut+\dfrac{9t^2}{2}$

O to D is

$ut+\dfrac{16t^2}{2}$

So AD is (OD-OA)=

$\dfrac{16t^2}{2}+ut$

$-\dfrac{t^2}{2}-ut$

$=\dfrac{15t^2}{2}$

And BC is (OC-OB) =

$\dfrac{9t^2}{2}+ut$

$-\dfrac{4t^2}{2}-ut$

$=\dfrac{5t^2}{2}$

Therefore, AD=3BC

Hence option C is correct.

A ball is released from the top of a tower of height

$h\ m$ . It takes

$T$ seconds to reach the ground. What is the position of the ball in

$\dfrac{T}{3}$ second?

0%
$\dfrac{h}{9}$ metres from the ground
0%
$\dfrac{7h}{9}$ metres from the ground
0%
$\dfrac{8h}{9}$ metres from the ground
0%
$\dfrac{17h}{18}$ metres from the ground

Explanation

Apply second equTION OF MOTION,

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

$h$ : height of the tower

$u = 0$ ; Initial velocity

$h = \dfrac{1}{2} gT^{2}$

$\dfrac{T}{3}\ s$ ,

$h_{1} = \dfrac{1}{2} g(\dfrac{T}{3})^{2}$

$\Rightarrow h_1 = \dfrac{1}{9}(\dfrac{1}{2} gT^{2})$

$\Rightarrow h_{1} = \dfrac{h}{9}$ from top.

Height from ground-

$h_g=h- h_{1} = h-\dfrac{h}{9}$

$\Rightarrow h_g = \dfrac{8h}{9}$ from ground

Two bodies, A(of mass 1kg) and B(of mass 3kg),are dropped from heights of 16 m and 25 m respectively. The ratio of the time taken by them to reach the ground is:-

0%
$\dfrac{5} {4}$
0%
$\dfrac{12} {5}$
0%
$\dfrac{5} {12}$
0%
$\dfrac{4} {5}$

Explanation

from equation of motion we know that when a body is dropped

$h = \dfrac{1}{2}gt^2$

$\dfrac{t_1}{t_2} = \sqrt{\dfrac{h_1}{1_2}} = \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}$

Hence (D) is correct answer

A particle starts with a velocity

$200\ cm/s$ and moves in a straight line with a retardation of

$10\ cm/s^{2}$ . Its displacement will be

$1500\ cm$ :

0%
Only once at $30\ s$ from start
0%
Only once at $10\ s$
0%
Twice at $10\ s$ and $30\ s$
0%
Always

Explanation

Given,

Initial velocity,

$u = 200\ cm/s$

Acceleration,

$a = - 10\ cm/s^{2}$

Dispplacement,

$s = 1500\ cm$

Using Newton's second equation of motion,

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

$1500 = 200t - \dfrac 12 \times 10 t^{2}$

$1500 = 200t - 5t^{2}$

$300 = 40t - t^{2}$

$t^{2}-40t +300 =0$

$t^{2}-10t-30t +300 =0$

$t(t-10)-30(t -10) =0$

$(t-10)(t-30)=0$

$\implies t =10\ s\ \text{and}\ 30\ s$

The displacement-time graph of motion of a particle is shown in the figure. The ratio of the magnitudes of the speeds during the first two seconds and the next four seconds is:

0%
$1:1$
0%
$1:2$
0%
$2:1$
0%
$1:\sqrt{2}$

Explanation

Slope of displacement-time curve gives the speed.
For first 2 seconds, speed

$v= \cfrac {40}{2} =20\ m/s$
For the last 4 seconds, speed

$v= \cfrac {40}{4} =10\ m/s$
Hence, ratio

$= 20/10=2:1$

A wooden block of mass

$10$ gm is dropped from the top of a cliff

$100$ m high. Simultaneously a bullet of mass

$10$ gm is fired from the foot of the cliff upward with a velocity

$100$ m/s. The bullet and the wooden block will meet each other after a time of:

0%
10 s
0%
0.5 s
0%
1 s
0%
7 s

A ball is released from the top of a tower of height

$h$ m. It takes

$T$ s to reach the ground. What is the position of the ball in

$\dfrac{T}{3}$ s?

0%
$\dfrac{h}{9}$ m from the ground
0%
$\dfrac{7h}{9}$ m from the ground
0%
$\dfrac{8h}{9}$ m from the ground
0%
$\dfrac{17h}{18}$ m from the ground

Explanation

Lets consider the equation

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

$u=0 , a=g$
So

$S=\dfrac{1}{2}gt^2$
Now

$h=\dfrac{1}{2}gT^2$
The distance traveled by ball in

$\dfrac{T}{3}$ s will be

$S=\dfrac{1}{2}g \times {(\dfrac{T}{3})}^2=\dfrac{h}{9}$
So height of ball above ground = (Total Height)

$-$ (Distance Traveled in

$\dfrac{T}{3}$ s)

$=h-\dfrac{h}{9}=\dfrac{8h}{9}$ m above ground

Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of

$10\ ms^{-1}$ . It implies that the boy is :

0%
at rest
0%
moving with no acceleration
0%
in accelerated motion
0%
moving with uniform velocity

A ball is thrown upwards. It takes 4 s to reach back to the ground. Find its initial velocity.

0%
$30 ms^{-1}$
0%
$10 ms^{-1}$
0%
$40 ms^{-1}$
0%
$20 ms^{-1}$

Explanation

$\text{Using}\ v=u+at$

Time of Flight

$= \dfrac{2u}{g}$
Where u is the initial velocity and g is acceleration due to gravity

$4=\dfrac{2u}{10}$

$u = 20 \ ms^{-1}$

A boy standing at the top of a tower of 20 $m$ height drops a stone. Assuming $g=10 ms^{-2}$ , the velocity with which it hits the ground is:

0%
$20 ms^{-1}$
0%
$40 ms^{-1}$
0%
$5 ms^{-1}$
0%
$10 ms^{-1}$

Explanation

When there is a free fall, we can directly use the equation of motion :

$v^{2} = u^{2} + 2gh$
here

$u = 0$
so,

$v = \sqrt{2gh}$ =

$\sqrt{2 \times 10 \times 20} = 20\ m/s$

A body is thrown vertically upwards with a velocity

$u$ , the greatest height

$h$ to which it will rise is:

0%
$u/g$
0%
$u^{2}/2g$
0%
$u^{2}/g$
0%
$u/2g$

Explanation

The body that is thrown vertically upward with velocity

$u$ will have final velocity

$v=0$ at the greatest height

$h$ .
Substituting the given values in the third equation of motion,

$v^2=u^2+2as$
we have

$0=u^2-2gh$ . (taking

$g$ in the upward direction)
or

$h=\dfrac{u^{2}}{2g}$ .

The velocity of a bullet is reduced from $100 \ m/s$ to $0 \ m/s$ while travelling through a wooden block of thickness $10\ cm$ . The retardation, assuming it to be uniform will be:

0%
$5 \times {10^4}\ m/{s^2}$
0%
$12 \times {10^4}\ m/{s^2}$
0%
$14 \times {10^4}\ m/{s^2}$
0%
$1 \times {10^4}\ m/{s^2}$

Explanation

$v^2 = u^2 +2as$

$\text{Substituting}\ v=0, u =100 m/s, s= 10\ cm = 0.1 \ m$

$a = - \dfrac {100^2}{0.1\times 2 }$

$a=- 5 \times 10^4 \ m/s^2$

Two objects of masses

$m_{1}$ and

$m_{2}$ having the same size are dropped simultaneously from heights

$h_{1}$ and

$h_{2}$ respectively. Find out the ratio of time they would take in reaching the ground.

0%
$\sqrt{\dfrac{h_{1}}{h_{2}}}.$
0%
$\sqrt{\dfrac{h_{2}}{h_{1}}}.$
0%
${\dfrac{h_{1}}{h_{2}}}.$
0%
${\dfrac{h_{2}}{h_{1}}}.$

Explanation

Gravitational acceleration of the objects will not depend on their mass or size and hence it will remain same for all the objects.

$h_1=u_1t_1+\dfrac{1}{2}at_1^2$

$h_2=u_2t_2+\dfrac{1}{2}at_2^2$

$u_1=u_2=0$

$\dfrac{t_1}{t_2} = (\dfrac {h_1}{h_2})^{\dfrac {1}{2}}$

A particle is moving in a circular path of radius

$r$ . The magnitude of displacement after half a circle would be :

0%
Zero
0%
$\pi r$
0%
$2 r$
0%
$2 \pi r$

A ball is dropped from height $h$ and another from $2h$ . The ratio of time taken by the two balls to reach ground is:

0%
$1:\sqrt{2}$
0%
$\sqrt{2}:1$
0%
$2:1$
0%
$1:2$

Explanation

Using equation of motion

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

$s = h, u = 0$

$\therefore t = \sqrt{\dfrac{2h}{g}}$
So, ratio of time taken:

$\dfrac{t_1}{t_2}=\sqrt{\dfrac{h_1}{h_2}}=\sqrt{\dfrac{h}{2h}}$

$=$

$\dfrac{1}{\sqrt 2}$

A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is

$S_1$ and that covered in the first 20 seconds is

$S_2$ then:

0%
$S_2$ = $S_1$
0%
$S_2$ = $2S_1$
0%
$S_2$ = $3S_1$
0%
$S_2$ = $4S_1$

If a body is thrown up with the velocity of $15 m/s$ , then maximum height attained by the body is $(g=10 \ m/s^2)$

0%
$11.25 m$
0%
$16.2 m$
0%
$24.5 m$
0%
$7.62 m$

Explanation

$v^2=u^2+2as$

$\text{Substituting}\ u =15\ m/s, v=0\ m/s, a=g=-10m/s^2,$

$15^2=2\times 10\times s$

$s=225/20=11.25m$

An electron starting from rest has a velocity that increases linearly with time,i.e. v=kt where

$k=2\ m/s^2$ . The distance covered in the first three seconds will be

0%
36m
0%
27m
0%
18m
0%
9m

Explanation

From the given information , v

$= kt$
Acceleration of the body 'a'

$= k = 2m/s^2$
Using the equations of motion

$s = ut + \frac{1}{2}at^2$
u = 0

$s = 2 \times (1/2) \times 3^2$

$s = 9m$

The velocity of a body at any instant is 10 m/s. After 5 sec, velocity of the particle is 20 m/s. The velocity at 3 seconds before is

(assume uniform accelaration)

0%
8 m/sec
0%
4 m/sec
0%
6 m/sec
0%
7 m/sec

Explanation

$v_i = 10\ m/s$ and

$v_f = 20\ m/s$
Acceleration of the body

$= \dfrac{20 - 10}{5} m/s^2$

$a = 2\ m/s^2$
Velocity of the body at

$t= -3\ sec$

$v = u + at$

$v = 10 - 2 \times 3$

$v = 4\ m/s$

In the velocity-time graph, AB shows that the body has

0%
uniform acceleration
0%
non-uniform retardation
0%
uniform speed
0%
initial velocity OA and is moving with uniform retardation

Explanation

$\textbf{Hint: }$

The slope of the curve in a velocity – time graph gives the acceleration.

$\textbf{Step1:Slope of velocity-time graph}$

Slope of velocity – time graph gives acceleration as shown below:

$slope=\dfrac{\Delta v}{\Delta t}=a$

$\textbf{Step2:Conclusion from graph}$

According to the graph of the question, the initial velocity of the body at time $t=0$ is $OA$ . The slope of the graph is constant but negative.

Acceleration is a vector quantity having both magnitude and direction. Increase in velocity with respect to time is called acceleration whereas decrease in velocity with respect to time is called retardation. In this graph, the velocity is decreasing at a constant rate with respect to time. Hence, the body shows uniform retardation.

The correct answer is option (D).

In the graph provided, the velocity

0%
increases between point 0 and A
0%
increases between point A and B
0%
decreases between points A and B
0%
is zero throughout

$N/kg$ is the unit of :

0%
Retardation
0%
Acceleration
0%
Rate of change of velocity
0%
All of these

Explanation

Acceleration or retardation

$a = \dfrac{Force}{Mass}$

Thus

$N \ kg^{-1}$ is the unit of retardation as well as acceleration.

Also, acceleration is defined as the rate of change of velocity.

Thus

$N \ kg^{-1}$ is also the unit of rate of change velocity.

Hence Option D.

A girl walks along a straight path to drop a letter in the letterbox and
comes back to her initial position. Her displacement-time graph is shown in the figure. Find the average velocity.

0%
1 m/s
0%
2 m/s
0%
-2 m/s
0%
0 m/s

Explanation

$\text{Average velocity} = \dfrac{\text{Total Displacement}}{\text{Total Time}}$

Since her displacement is

$zero$

Average velocity

$=0/100=0\ m/s$

The rate of change of displacement with time is

0%
speed
0%
acceleration
0%
retardation
0%
velocity

Explanation

Correct answer: Option D

Explanation:

Velocity is defined as the rate at which an object's displacement changes over time (displacement over time). Because it has both magnitude (called speed) and direction, velocity is a vector. It is written as,

$velocity = \dfrac{{Change{\text{ }}in{\text{ }}displacement}}{{time}}$

Hence, The rate of change of displacement with time is velocity

Rain is falling vertically with a speed of 1.7 m/s. A girl is walking with speed of 1.0 m/s in the N - E (north-east) direction. To shield herself she holds her umbrella making an approximate angle

$\theta$ with the vertical in a certain direction. Then

0%
$\theta=60^0 \:in \:N - E \:direction$
0%
$\theta=30^0 \:in \:N - E \:direction$
0%
$\theta=60^0 \:in \:S - W \:direction$
0%
$\theta=30^0 \:in \:S - W \:direction$

A particle is moving in a circular path of radius r. Its displacement after moving through half the circle would be :

0%
Zero
0%
$r$
0%
$2r$
0%
$\dfrac{2}{r}$

Explanation

Displacement is the minimum distance between the initial position and final position.
After moving half circle (A to B) displacement will be AB
AB is diameter i.e. $2r$
Option C is correct.

Two balls are dropped from height

$h_1$ and

$h_2$ respectively. What is the ratio of their velocities on reaching the ground is?

0%
$\sqrt{\dfrac{h_2}{h_1}}$
0%
$\sqrt{\dfrac{h_1}{h_2}}$
0%
$\dfrac{h_2}{h_1}$
0%
$\dfrac{h_1}{h_2}$

Explanation

From 3rd equation of motion,

$v^2=u^2+2as$

Given

$u=0$ for both the balls since they are dropped.

Hence,

$v \propto \sqrt{s}$

$\dfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\dfrac { \sqrt { { h }_{ 1 } } }{ \sqrt { { h }_{ 2 } } }$

A body starting from rest in travelling with an acceleration of

$6m/s^2$ . Find the distance traveled by it in

$6^{th}$ second.

0%
$36$ mts
0%
$28$ mts
0%
$54$ mts
0%
$33$ mts

Explanation

Given,

$n=6$

$u=0m/s$

$a=6m/s^2$

The distance travelled by the body in

$n^{th}$ second

$S_n=u+\dfrac{a}{2}(2n-1)$

$s_6=0+\dfrac{6}{2}(2\times 6-1)=33mts$

The correct option is D.

From kinematics, correct equation of motion for

$S$ is/are :

0%
$S=ut-\frac {1}{2}at^2$
0%
$S=ut+\frac {1}{2}at^2$
0%
Both (a) and (b)
0%
None of these

Explanation

According to the second equation of motion,

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

where,

$s$ =displacement

$u$ =initial velocity

$t$ = time

$a$ =acceleration

(i) If the displacement and acceleration are in the same direction, then,

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

where

$a$ is the magnitude of the acceleration

For example, a car linearly accelerating on road.

(ii) If the displacement and acceleration are in the opposite direction, then,

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

where

$a$ is the magnitude of the acceleration

For example, a ball thrown vertically upwards in air

Hence, the correct option is C.

Speed of a body depends on

0%
Time
0%
Mass of the body
0%
Distance travelled by the body
0%
Both A and C

Explanation

Speed is defined as

$Speed = \dfrac{\text{Distance travelled}}{\text{Time interval}}$

Hence speed depends both on time and distance travelled by the body.

So both A and C are correct.

A particle is projected up with a velocity of

$\sqrt{29}\ ms^{-1}$ from a tower of height 10m. Its velocity on reaching the ground is

$x$

$\ ms^{-1}$ . Find

$x$ .

0%
5
0%
10
0%
15
0%
20

Average speed of a body is

0%
Total Distance $\div$ Total Time
0%
Total Distance + Total Time
0%
Both (a) and (b)
0%
None of these

Explanation

We know that,

The average speed of a body is the ratio of total distance and total time.

So, Average speed =

$\dfrac{\text{Total distance}}{\text{Total time}}$

$v_{av}=\dfrac{D}{T}$

Here,

$D$ = Total distance

$T$ = Total time

Hence, Option A is the correct.

If a body is thrown up with an initial velocity u and covers a maximum height of h, then h is equal to

0%
$\dfrac{u^2}{2g}$
0%
$\dfrac{u}{2g}$
0%
$2u^2g$
0%
None of these

Explanation

At the maximum height , velocity of the ball is zero

$v^2 = u^2 - 2gh$
v

$= 0$

$u^2 = 2gh$

$h = \dfrac{u^2}{2g}$

A stone is projected up with a velocity of

$4.9\ m/s$ from the top of a tower and it reaches the ground after 3 s. Then the height of that tower is

0%
4.9 m
0%
19.6 m
0%
9.8 m
0%
29.4 m

Explanation

From second equation of motion,

$s=ut+at^2/2$

Assume upward direction to be positive.

$u=4.9\ m/s,\ t=3\ sec,\ a=-9.8\ m/s^2$

$s=4.9\times 3-9.8\times 3^2/2$

$s =-29.4\ m$

Height = 29.4 m

If an object is thrown vertically up with a velocity of

$19.6\ ms^{-1}$ , it strikes the ground after

$x$ second, find

$x$ .

0%
1
0%
2
0%
4
0%
8

Explanation

$u=19.6m/s$

Total time taken

$T=\dfrac{2u}{g}$

$x=T=2\times \dfrac{19.6}{9.8}$

$x=4sec$

1319295_100907_ans_a30505b650d94675bf7e9273da91a026.png

The relationship between average speed, time and distance is

0%
Average speed = distance $\times$ time
0%
Average speed $= \cfrac{\text{total distance}}{\text{total time}}$
0%
$\text{Time} = \cfrac{\text{average speed}}{\text{distance}}$
0%
$\text{Distance} = \text{average speed } \times \text{time}$

A freely falling body from rest acquires velocity V falling through a distance h. After the body falls through a further distance h velocity acquired by it is

0%
2v
0%
v
0%
$\sqrt{2}$ v
0%
4v

Explanation

Using the equations of motion

$v^2 = u^2 + 2gh$ (u is 0)

$v^2 = 2gh$

$V = \sqrt{2gh}$

$v^2 = u^2 + 2gh_2$

$V_2^2 = 4gh$

$V_2^2 = 4gh = 2V^2$

$V_2 = \sqrt{2} V$

An object is thrown vertically upward with a velocity of

$10\ ms^{-1}$ . It strikes the ground after _______ seconds. (Take

$g = 10\ ms^{-2}$ )

0%
10
0%
2
0%
5
0%
1

Explanation

Given:

Initial velocity

$u=10\ m/s$

Gravitational acceleration

$g=10\ m/s^2$

$\text{Time to ascend maximum height}$ :

When the object is ascending,

$g$ is causing deceleration. At the maximum height, the velocity of the object is zero.

Let

$t_1$ is the time required to reach the maximum height.

Thus, using the equation

$v=u-gt$ we get,

$0=10-(10\times t_1)$

$\Rightarrow t_1=1\ sec$

$\text{Time required to reach the maximum height} = \text{Time required to fall down to the ground from the maximum height}$

Thus, the total time required is

$(1+1)=2\ sec$

A stone is dropped from a rising balloon at a height of 300 m above the ground and it reaches the ground in 10 s. The velocity of the balloon when it was dropped is :

0%
$19 m s^{-1}$
0%
$19.6 m s^{-1}$
0%
$29 m s^{-1}$
0%
$0 m s^{-1}$

Explanation

$h = -ut + \displaystyle \dfrac{1}{2} gt^2$

$300 = - 10 u \displaystyle + \dfrac{1}{2} \times 9.8 \times 100$

$-10 u = 300 - 490 = -190$

$u \displaystyle = \dfrac{190}{10}= 19 m s^{-1}$

An object is released from a balloon rising up with a constant speed of

$2\ ms^{-1}$ . Its magnitude of velocity after

$1\ s$ in

$\ ms^{-1}$ is

0%
5.3
0%
7.8
0%
2.8
0%
6.4

Explanation

The ball initially will have the same velocity as that of the balloon
Using the equations of motion , taking velocity to be positive upwards and negative downwards

$v = u + at$

$v = 2 - g(1)$

$v = 7.8 m/s$

When a body is projected vertically up from the ground its velocity is reduced to

$\frac{1}{4}$ th of its velocity at ground at height

$h$ . Then the maximum height reached by the body is

0%
$\dfrac{15}{32}h$
0%
$\dfrac{15}{16}h$
0%
$\dfrac{15}{8}h$
0%
$\dfrac{5}{4}h$

Explanation

Using the equation of motion ,

$v^2 = u^2 - 2gh$

$\dfrac{u^2}{16} = u^2 - 2gh$

$h = \dfrac{15 u^2}{32 g}$

At Max. height,

$v=0$ and equation of motion is

$v^2 = u^2 - 2gh$

$H = \dfrac{u^2}{2g} =\dfrac{15h}{16}$

A car travels with a uniform velocity of

$25\, m\, s^{-1}$ for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s. Find the acceleration.

0%
$5\, m\, s^{-2}$
0%
$-2.5\, m\, s^{-2}$
0%
$25\, m\, s^{-2}$
0%
$10\, m\, s^{-2}$

Explanation

For retardation process,

$u = 25 m/s, v=0, t=10 s$

$v=u+at$

$0=25+a\times 10$

$a = -2.5 m/s^2$

A body projected vertically up with a velocity of 10 m s

$^{-1}$ reaches a height of 20 m. If it is projected with a velocity of 20 m s

$^{-1}$ , then the maximum height reached by the body is :

0%
20 m
0%
10 m
0%
80 m
0%
40 m

Explanation

$u = 10 m s^{-1} ; h = 20 m$

$u_1 = 20 m s^{-1}; h_1 = ?$

Formula for maximum height

$h=\frac{u^2}{2g}$

hence

$\displaystyle \dfrac{h}{h_1}= \dfrac{u^2}{u_1^2} ; \dfrac{20}{h_1} = \dfrac{10^2}{20^2}$

$h_1 \displaystyle = \dfrac{20 \times 400}{100} = 80 m$

CBSE Questions for Class 9 Physics Motion Quiz 4 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Motion Quiz 4 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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