MCQ Questions for Class 9 Physics Motion Quiz 3

The graph above shows the position versus time for three different cars 1,2, andRank these cars according to the magnitudes of their velocities, greatest first.

0%
1, 2, 3
0%
3, 2, 1
0%
2, 1, 3
0%
2, 3, 1
0%
all tie

A ball of mass

$m$ is thrown straight upward from the top of a multi-storey building with an initial velocity of

$15 \ m/s$ .
Find out the time taken by the ball to reach its maximum height?

0%
$25 \ s$
0%
$15 \ s$
0%
$10 \ s$
0%
$5 \ s$
0%
$1.5 \ s$

Explanation

Given : initial velocity

$u = 15 \ m/s$

acceleration

$a = -g = -10 m/s^2$ (taking upward direction to be positive)

Final velocity of the ball at the highest point

$v = 0 \ m/s$

Using the first equation of motion

$v = u + at$

$\therefore$

$0 = 15 + (-10) t$

$\implies t = 1.5 \ s$

A car accelerates from rest at a rate of

$4 m/s^2$ for

$10 \ s$ . Calculate the speed of the car at the end of

$10 \ s$ ?
[Unless otherwise stated, use

$g=10m/s^2$ and neglect air resistance ]

0%
14 m/s
0%
6 m/s
0%
2.5 m/s
0%
0.4 m/s
0%
40 m/s

Explanation

Given :

Initial velocity

$u = 0$ m/s

Acceleration

$a = 4$

$m/s^2$

Time

$t = 10$ s

Using first equation of motion

$v = u+at$

$\therefore$

$v = 0 + 4 (10) = 40 \ m/s$

The above graph shows position as a function of time for an object moving along a straight line. During which time(s) is the object at rest?
I.

$0.5$ seconds
II. From

$1$ to

$2$ seconds
III.

$2.5$ seconds

0%
I only
0%
II only
0%
III only
0%
I and III only

Explanation

Slope of the curve under position-time graph gives the instantaneous velocity of the object.

Slope of the curve is zero only in the time interval

$1<t<2$ s.

Thus the object is at rest (or velocity is zero) only from

$1$ to

$2$ s.

Hence option B is correct.

Match the entries in Column-I with those in Column-II.

	Column - I		Column - II
a	Speed	1	cm
b	Acceleration	2	s
c	Displacement	3	$m s^{-1}$
d	Time	4	$km h^{-2}$

0%
a-2, b-1, c-4, d-3
0%
a-3, b-4, c-1, d-2
0%
a-4, b-2, c-3, d-1
0%
a-3, b-2, c-2, d-1

Explanation

In a given table, Column-I has given Physical quantity and Column-II has given units.

Displacement: Minimum distance between two points and its C.G.S unit is

$cm.$

Speed: It is defined as the ratio of distance over time and its S.I unit is

$m/s.$

Time: It is defined as the duration in which all things happen and it unit is seconds (s).

Acceleration: it is equal to velocity over time and its S.I unit is

$m/s^2$ , its unit can also be written as

$kmh^{-2}$

So correct match is,

Column-I	Column-II
(a) Speed	(3) $m/s$
(b) Acceleration	(4) $kmh^{-2}$
(c) Displacement	(1) $cm.$
(d) Time	(2) $s$

A football player initially at rest starts 20 meters from the goal line. He accelerates towards the goal line at 2m/s

$^2$ . Find out the distance between player and goal line after 4s?

0%
8 m
0%
28 m
0%
32 m
0%
4 m
0%
52 m

Explanation

Given :

$u = 0$ m/s

$t =4$ s

$a =2 m/s^2$

Using

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

Thus distance covered in 4 s

$S = 0 + \dfrac{1}{2} \times 2 \times 4^2 = 16$ m

Thus distance between player and goal

$d =20 -16 = 4$ m

In a uniform circular motion, the magnitude and direction of velocity at different points remain the same.

0%
True
0%
False

The accelerated motion of a body changes due to change in:

0%
Speed
0%
Direction of motion.
0%
Velocity.
0%
All of the above

Uniform circular motion is called continuously accelerated motion mainly because its :

0%
direction of motion changes
0%
speed remains the same
0%
velocity remains the same
0%
direction of motion does not change

A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 12 m

$s^{-1}$ , and decreases at the rate of 0.5 m

$s^{-2}$ . Find the time it will take to come to rest.

0%
6 s
0%
12 s
0%
24 s
0%
48 s

Explanation

$v=u+at$

$\therefore t=\dfrac{v-u}{a}$

$v=0, u=12ms^{-1}, a=-0.5ms^{-2}$

$t=\dfrac{0-12}{-0.05}=24s$

A ball is thrown vertically upward with a speed of

$25.0\ m/s$ . How high does it rise?

0%
$31.9 m$
0%
$32 m$
0%
$33 m$
0%
$33.9$

Explanation

Given :

$u=25m/s$ (Taking upward direction to be positive)

$v=0m/s$

$a=-g = -9.8 \ m/s^2$

$S=H$
Applying

$3rd$

$eqn$ of motion,

$v^{2}=u^{2}+2aS$

$0^{2}=25^{2}+2 \times (-g) \times H$

$H= \dfrac{625}{2\times 9.8}$

$H=31.9m/s$

A truck and a car are moving with equal velocity. On applying brakes, if both decelerate at the same rate,what happens?

0%
The truck will cover less distance before stopping.
0%
The car will cover less distance before stopping.
0%
Both will cover equal distance.
0%
None of the above

Explanation

Consider, the car has a mass ‘m’ and truck has a mass ‘M’.

where ,

$M>>m$

Momentum = (mass of the body) x (velocity of the same body)

As, the velocities of the car and truck are the same….its the mass which decides the distance which will be covered after applying brakes by each of them.

As the mass of truck is much higher than that of a car, the truck has a larger momentum value.

HENCE, THE TRUCK TRAVELS A LARGER DISTANCE AFTER APPLYING BRAKES.

A car starts from rest and accelerates uniformly over a time of

$5.21$ seconds for a distance of

$110$ m. Determine the acceleration of the car.

0%
$8.10 m/s/s$
0%
$8.1 m/s/s$
0%
$8.0 m/s/s$
0%
$7.10 m/s/s$

Explanation

Time taken

$t=5.21 s$

Initial velocity

$u=0$ and distance traveled

$s=110 m$

Using,

$s=ut+at^2/2$

where

$a$ is the acceleration.

$\Rightarrow 110=0(5.21)+a(5.21)^2/2$

We get

$a=8.10 m/s^2$

Choose the correct option.

Odometer: Distance :: ______:______

0%
Weighing scale : Weight
0%
Length : Width
0%
Inch : Foot
0%
Mileage : Speed

In circular motion, the

0%
Direction of motion is fixed
0%
Direction of motion changes continuously
0%
Acceleration is zero
0%
Velocity is constant

Explanation

$\textbf{Explanation:}$

The speed of the ball is constant. It traces a circle with a fixed center.

The body forms a circle by moving in a rounded path. With respect of time, the body's direction is always changing. Along the rounded path, the body moves in a tangential motion.
Hence option B is correct.

In a uniform linear motion which of the following quantities remains zero?

0%
Speed
0%
Velocity
0%
Acceleration
0%
None of these

Explanation

In uniform linear motion, velocity is constant which means acceleration is zero.

$\text{acceleration} = \dfrac{ \text{ change in velocity } }{ \text{ time } }$

An uniform circular motion is an uniform velocity motion

0%
True
0%
False

Speed can be defined as :

0%
$\dfrac{Distance }{time}$
0%
$\dfrac{Distance }{{Time}^2}$
0%
$\dfrac{Time}{Distance}$
0%
$None\ of\ the\ above$

Explanation

Speed of the body is defined as the distance travelled by the body in unit time.
Mathematically,

$Speed = \dfrac{Distance}{time}$

If a particle moves in a closed path with a constant speed of 10 m/s, we can always say, that there will be a change in velocity.

0%
True
0%
False

The velocity of a body moving with a uniform acceleration of

$2 m/sec^2$ is

$10 m/sec$ . Its velocity after an interval of 4 sec is

0%
12 m/sec
0%
14 m/sec
0%
16 m/sec
0%
18 m/sec

Explanation

By the first eqn of motion

$v=u+at$ Put the given values

$v=10+2\times 4$

$u=18\ m/sec$

The rate of change of velocity is:

0%
Force
0%
Momentum
0%
Acceleration
0%
Displacement

A body moving along a straight line at 20 metre per second undergoes an acceleration of 4 metre per second square. After 2 second it speed will be:

0%
12 metre per second
0%
28 metre per second
0%
72 metre per second
0%
20 metre per second

Ball A is dropped freely while another ball B is thrown vertically downward with an initial velocity

$v$ from the same point simultaneously. After

$t\ seconds$ they are separated by a distance of: (Given that

$t$ is less than the time required by ball B to reach the ground.)

0%
$\dfrac{vt}{2}$
0%
$\dfrac{1}{2}gt^{2}$
0%
$vt$
0%
$vt+\dfrac{1}{2}gt^{2}$

Explanation

First case (for ball A):

$s_{1}=0(t)+\dfrac{1}{2}gt^{2}$
Second case (for ball B):

$s_{2}=v(t)+\dfrac{1}{2}gt^{2}$
after

$t\ seconds$ , distance between them is

$s_{2}-s_{1}$

$\Rightarrow s_{2}-s_{1}=vt$

Note:

$\text{After ball B has landed on ground the distance between A and B decreases and becomes zero.}$

So, C is correct.

A body starting with a velocity

$v$ returns to its initial position after

$t$ second with the same speed along the same line. Acceleration of the particle is :

0%
$\dfrac{-2v}{t}$
0%
$\dfrac{2v}{t}$
0%
$\dfrac{v}{2t}$
0%
$\dfrac{t}{2v}$

Explanation

$\textbf{Step 1 - Calculating acceleration Refer Figure}$

Given,

Initial velocity

$= v$

Final velocity

$= -v$ and

$time = t$

$v = u + at$

$-v = v + at$

$\Rightarrow a = \dfrac {-2v}{t}$

Correct option : A.

A body dropped from the top of a tower reaches the ground in

$4\;s$ . Height of the tower is (Take

$g=9.8 \ m/s^2$ ):

0%
$39.2\ m$
0%
$44.1\ m$
0%
$58.8\ m$
0%
$78.4\ m$

Explanation

We have,

$s = ut + \frac {1}{2} at^2$
Given,

$t = 4\ sec$

$a = g$

$u = 0$

$\Rightarrow s = \dfrac {1}{2} gt^2$

$s = \dfrac {1}{2} g(4)^2 = 8g = 8 \times 9.8 = 78.4\ m$
Height of the tower = 78.4 m

If the positive direction is downward, which situation involves negative velocity and zero acceleration?

0%
A rocket slowing down as it moves upward
0%
A rocket moving downward at a constant speed
0%
A rocket moving upward at a constant speed
0%
A rocket speeding up as it moves upward

The displacement - time graphs of two bodies A and B are OP and OQ respectively. If

$\angle$ POX is

$60^{0}$ and

$\angle$ QOX is

$45^{0}$ , the ratio of the velocity of A to that of B is

0%
$\sqrt{3}:\sqrt{2}$
0%
$\sqrt{3}:1$
0%
$1: \sqrt{3}$
0%
$3:1$

Explanation

$V_{A}= \tan 60^{0}$

$V_{B}= \tan 45^{0}$

So,

$\dfrac{V_{A}}{V_{B}} = \dfrac{\tan 60^o}{\tan 45^o} = \dfrac{\sqrt{3}}{1} = \sqrt{3}: 1$

A body falls freely from rest. If at an instant, the velocity acquired is numerically equal to the displacement, then the velocity acquired is:

0%
$9.8\ m/s$
0%
$19.6\ m/s$
0%
$29.4\ m/s$
0%
$39.2\ m/s$

Explanation

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

$a = g$

$|v| = |s|$

$u = 0$

$v^2 = 0^2 + 2gv$

$v = 2g$

$|v| = 2g = 2 \times 9.8 = 19.6\ m/s^2$

The figure shows four graphs of displacement (

$x$ ) versus time (

$t$ ), the graph that shows a constant, positive, and non-zero velocity is:

0%
$(a)$
0%
$(b)$
0%
$(c)$
0%
$(d)$

Explanation

When an object travels equal distances in equal intervals of time, it moves with uniform speed. This shows that the distance traveled by the object is directly proportional to the time taken. Thus, for uniform speed, a graph of distance traveled against time is a straight line.

$Graph\ (a)-$ The graph is a straight line and displacement is increasing with increasing time. Therefore, velocity is uniform (constant), increasing (positive), and non-zero.

$Graph\ (b)-$ The graph is a straight line but displacement is constant with increasing time. Therefore, velocity is zero.

$Graph\ (a)-$ The graph is a not straight line. Therefore, velocity is non-uniform and non-zero.

$Graph\ (d)-$ The graph is a straight line and displacement is decreasing with increasing time. Therefore, velocity is uniform (constant), decreasing (negaitive), and non-zero.

A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?

0%
x < 0, v < 0, a > 0
0%
x > 0, v < 0, a < 0
0%
x > 0, v < 0, a > 0
0%
x > 0, v > 0, a < 0

If north is the positive direction, which situation involves positive velocity and negative acceleration?

0%
A car speeding up as it moves southward
0%
A car moving southward at a constant speed
0%
A car slowing down as it moves northward
0%
A car slowing down as it moves southward

The distances travelled by a body starting from rest and travelling with uniform acceleration in successive intervals of time each of one second will be in the ratio:

0%
$1:2:3$
0%
$1:2:4$
0%
$1:3:5$
0%
$1:5:9$

Explanation

Distance traveled (s) is given by :

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

Now, interval wise distance traveled are:

$s_1 - s_0 = \dfrac {a}{2} (1)^2 = \dfrac {a}{2}$

$s_2 - s_1 = \dfrac {a}{2} (2)^2 - \dfrac {a}{2} (1)^2 = \dfrac {3a}{2}$

$s_3 - s_2 = \dfrac {a}{2} (3)^2 - \dfrac {a}{2} (2)^2 = \dfrac {5a}{2}$

Hence, its in the ratio

$= \dfrac {a}{2} : \dfrac {3a}{2} : \dfrac {5a}{2}$

$= 1 : 3 : 5$

A body reaches the ground in

$3\;s$ when it is released with zero velocity from the top of a building. The height of the building is:

0%
$14.7\ m$
0%
$24.4\ m$
0%
$44.1\ m$
0%
$66.2\ m$

Explanation

Given,

Initial velocity,

$u=0$

Acceleration due to gravity,

$a=g=9.8\ m/s^2$

Time taken,

$t=3\ s$

Height of the building,

$h$

Using Newton's second equation of motion,

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

$s=0+\dfrac{1}{2}gt^2$

$s= \dfrac{1}{2}9.8 \times(3)^2$

$s= 44.1\ m$

The displacement time graph of a moving particle is shown in the figure. The instantaneous velocity of the particle is negative at the point:

0%
D
0%
F
0%
C
0%
E

A balloon starts from rest, moves vertically upwards with an acceleration

$\dfrac{g}{8}\ \text{ms}^{-2}$ . A stone falls from the balloon after

$8\ \text{s}$ from the start. The time taken by the stone to reach the ground is

$\left(g=9.8\ \text{ms}^{-2}\right)$ :

0%
$4\; s$
0%
$8\; s$
0%
$2$ $\sqrt{2}$ s
0%
$12\; s$

Explanation

For balloon, distance traveled by it in 8 sec,

$s = ut + \dfrac {1}{2} gt^{2}$ where $u=0$

$s$ $=$ $\dfrac{1}{2}\times \dfrac{g}{8} t^2$ $=4g$

So, time taken by stone,

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

$4g = \dfrac {1}{2} gt^2$ at this height the velocity of stone will be zero

$t^2 = \sqrt 8 \Rightarrow t = 2 \sqrt 2\ s.$

A pebble is thrown vertically upwards from a bridge with an initial velocity of

$4.9 \;ms^{-1}$ . It strikes the water after

$2\;s$ . Height of the bridge is:

0%
$19.6$ m
0%
$14.7$ m
0%
$9.8$ m
0%
$4.9$ m

Explanation

Option C is correct

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

$u = + 4.9 m/s$

$g = -9.8 m/s^2$

$t = 2 sec.$

$\Rightarrow s = 4.9 \times 2 - \dfrac {1}{2} \times 9.8 \times (2)^2 = -9.8 m$
Height is

$|s| = 9.8\ m$

A body thrown vertically upwards reaches the highest point in

$2\;s$ . Initial velocity of projection is_____

0%
9.8 $ms^{-1}$
0%
19.6 $ms^{-1}$
0%
29.4 $ms^{-1}$
0%
39.2 $ms^{-1}$

Explanation

$v = u + at; a = -g$

$v = u - gt$

$v = 0, u = gt$

$u = 9.8 \times 2 = 19.6\ m/s$

A body thrown up with some initial velocity reaches a maximum height of

$50\;m$ . Another body with double the mass thrown up with double the initial velocity will reach a maximum height of :

0%
$100$ m
0%
$200$ m
0%
$400$ m
0%
$50$ m

A body projected up with a velocity

$u$ reaches maximum height

$h$ . To reach double the height, it must be projected up with a velocity of:

0%
$2u$
0%
$\dfrac{u}{2}$
0%
$\sqrt{2}u$
0%
$\dfrac{u}{\sqrt{2}}$

Explanation

$\textbf{Step 1 - Calculating height Refer Figure}$

Given,

initial height,

$h_{1} = h$

final height,

$h_{2} = 2h$

As we know, at maximum height final velocity

$(v)$ become zero.

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

$0 = u^{2} - 2gh$

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

$....(1)$

Similarly,

$2h = \dfrac {(u_{2})^{2}}{2g}$

$....(2)$

$\textbf{Step 2 - Dividing equation (1) and (2)}$

$\dfrac {h}{2h} = \dfrac {u^{2}}{2g}\times \dfrac {2g}{(u_{2})^{2}}$

$\Rightarrow \dfrac {1}{2} = \dfrac {u^{2}}{(u_{2})^{2}}$

$\Rightarrow u_{2} = \sqrt {2}u$

Correct option : C.

A ball dropped freely takes

$0.2\;s$ to travel the last

$6\;m$ distance before hitting the ground. Total time of fall is

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

0%
$2.9\ s$
0%
$3.1\ s$
0%
$2.7\ s$
0%
$0.2\ s$

Explanation

Let

$T=$ total time taken in second

$S=$ total distance traveled in meter

$V=$ velocity of particle in m/sec
So in this question the condition is of free fall so acceleration of the particle will remain constant through out the process.

${T}={T}_{1}+{T}_{2}$

${S}={S}_{1}+{S}_{2}$
Given that:

${T}_{1}=0.2 \;s$ and

${S}_{1}=6 \;m$

Thus,

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

Thus, we get

$u=29 m/s$

Now using:

$u= 0 +gt$ (as the ball is dropped with initial velocity as zero)

$t=\dfrac{29}{10}=2.9 s$

Thus we get total time as

$2.9+0.2=3.1 s$

The average velocity of a body moving with uniform acceleration after travelling a distance of

$3.06\ m$ is

$0.34\ {m/s}$ . The change in velocity of the body is

$0.18\ {m/s}$ . During this time, its acceleration is

0%
$0.01\ {m/s}^{2}$
0%
$0.02\ {m/s}^{2}$
0%
$0.03\ {m/s}^{2}$
0%
$0.04\ {m/s}^{2}$

Explanation

$V_{avg} = \dfrac {D}{t} \Rightarrow 0.34 = \dfrac {3.06}{t}$

$\Rightarrow t = \dfrac {3.06}{0.34}$ =

$9\ sec$

We have,

$v = u + at$

$v - u = at$

$0.18 = a \times 9$

$\Rightarrow a = \dfrac {0.18}{9} = 0.02 \ m/s^2$

Two bodies are projected simultaneously with the same speed of

$19.6 \;ms^{-1}$ from the top of a tower, one vertically upwards and the other vertically downwards. As they reach the ground, the time gap is:

0%
0 s
0%
2 s
0%
4 s
0%
6 s

Two balls are dropped to the ground from different heights. One ball is dropped

$2\ s$ after the other but both of them strike the ground simultaneously

$5\ s$ after the first is dropped. The difference in the heights, when they are dropped is (

$g=10\ m/s^2$ )

0%
80 m
0%
45 m
0%
95 m
0%
100 m

Explanation

Ball B is dropped first. Then, ball A is dropped after 2 s.

$h_B = \dfrac {1}{2}gt^2 = \dfrac {1}{2}g(5)^2 = 12.5g = \dfrac {25g}{2}$ ....... (1)

$h_A = \dfrac {1}{2}gt^2 = \dfrac {1}{2}g(3)^2 = \dfrac {9g}{2}$ ........ (2)

$eq^n (1) - eq^n (2) \Rightarrow h_B - h_A = \dfrac {25g}{2} - \dfrac {9g}{2} = \dfrac {16g}{2} = \dfrac {16 \times 10}{2}=80\ m$

A body released from a height falls freely towards the earth. Another body is released from the same height exactly a second later. Then the separation between the two bodies two seconds after the release of the second body is

0%
4.9 m
0%
9.8 m
0%
19.6 m
0%
24.5 m

Explanation

Both the bodies are falling freely with acceleration

$g = 9.8\ m/s^2$ .

Equation to be used:

$s=\dfrac{1}{2}\times g\times t^2$

The first body descends for 3 sec.

The height covered by this body is

$s_1 = \dfrac {1}{2} \times (9.8) (3^2)=44.1\ m$

The second body descends for 2 sec.

The height covered by this body is

$s_2 = \dfrac {1}{2} \times (9.8) (2^2)=19.6\ m$

Separation between the bodies is

$s_1 - s_2 = (44.1-19.6) = 24.5\ m$

A body is thrown vertically upwards and rises to a height of 10 m. The velocity with which the body was thrown upwards is(

$g=9.8\ m/s^2$ )

0%
16 m/s
0%
15 m/s
0%
14 m/s
0%
12 m/s

Explanation

Using the formula

$v^2 = u^2 - 2gh$
At maximum height , v = 0

$u^2 = 2gH$

$u^2 = 2 \times 9.8 \times 10$

$u^2 = 196$

$u = 14 m/s$

A stone is dropped from a balloon at an altitude of

$280\ m$ . If the balloon ascends with a velocity of

$5\ ms^{-1}$ and descends with a velocity of

$5\ ms^{-1}$ , times taken by the stone to reach the ground in the two cases respectively are (

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

0%
$8\ s$ and $9\ s$
0%
$9\ s$ and $8\ s$
0%
$3\ s$ and $4\ s$
0%
$8\ s$ and $7\ s$

Explanation

Initial speed of the particle becomes same as the ballon have.

Assume upward direction as positive and downward direction as negative.

$1^{st}$ Case:

Ascend with

$5\ m/s$

$u = 5\ m/s$ : Initial velocity of particle

Apply second equation of motion.

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

$\Rightarrow s = 5t - 5t^2$

Displacement is

$-280\ m$ , in downward direction

$\Rightarrow 5t-5t^2= -280$

$\Rightarrow t = 8\ sec$

$2^{nd}$ Case:

Descend with

$5\ m/s$

$u = -5\ m/s$ : Initial velocity of particle

Apply second equation of motion.

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

Displacement is

$-280\ m$ , in downward direction

$\Rightarrow -280 = -5t - 5t^2$

$\Rightarrow t = 7\ sec$

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at

$2\ m/ s^{2}$ . He reaches the ground with a speed of

$3\ m/ s$ . At what height, did he bail out ? (Given

$g=9.8\ m/s^2$ approximately)

0%
91 m
0%
182 m
0%
293 m
0%
111 m

Explanation

The speed of the person just before the parachute opens is:

$u=\sqrt {2gh}=\sqrt {2\times 9.8\times 50}=\sqrt {980}$

When the parachute opens and descends,

$v^2=u^2+2(-a)h$

$v^2-u^2=-2(2)(h)$

$9-980=-4h$

$\Rightarrow h=971/4=242.75\ m$

$\therefore$ The total height of fall is

$242.75+50=292.75\ m\approx293\ m$

A body is thrown vertically up with certain velocity. If

$h$ is the maximum height reached by it, its position when its velocity reduces to

$(1/3)^{rd}$ of its velocity of projection is at

0%
8h/9 from the ground
0%
8h/9 below the top-most point
0%
4h/9 from the ground
0%
h/3 below the top-most point

Explanation

Let initial velocity is

$u$ ,

$h = \dfrac {u_1^2}{2g}$

$v^2 = u^2 - 2gx$

$(\dfrac {u_1}{3})^2 = u_1^2 - 2gx$

$2gx = \dfrac {8}{9} u_1^2$

$x = \dfrac {8}{9} (\dfrac {u_1^2}{2g}) = \dfrac {8}{9}h$

A ball is dropped from the top of a building. The ball takes

$0.5$ s to fall past the window

$3$ m in length at certain distance from the top of the building. (

$g=10\;ms^{-2}$ )
Speed of the ball as it crosses the top edge of the window is

0%
$3.5 \;ms^{-1}$
0%
$8.5 \;ms^{-1}$
0%
$5 \;ms^{-1}$
0%
$12\; ms^{-1}$

A person standing on the edge of a well throws a stone vertically upwards with an initial velocity

$5ms^{-1}$ . The stone goes up, comes down and falls in the well making a sound. If the person hears the sound 3 second after throwing, then the depth of water is (neglect time travel for the sound and take

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

0%
$1.25$ m
0%
$21.25$ m
0%
$30$ m
0%
$32.5$ m

Explanation

$h = ?$ Taking vertically downward direction as negative,

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

$t = 3 s$

$h = 5(3) - \dfrac {1}{2}g(3)^2 = 15-\dfrac {9g}{2}$

$= 15 - 45 = 30\ m$

Hence, depth

$= 30\ m$

CBSE Questions for Class 9 Physics Motion Quiz 3 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Motion Quiz 3 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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