# Applications of equations of motion

## Trending Questions

**Q.**A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height h2. The velocity versus height of the ball during its motion may be represented graphically by

(graph are drawn schematically and on not to scale)

**Q.**A ball is thrown vertically upwards with a speed v from a height h metre above the ground. The time taken for the ball to hit ground is

- vg√1−2hgv2
- vg√1+2hgv2
- √1+2hgv2
- vg[1+√1+2hgv2]

**Q.**Two particles 1 and 2 are allowed to descend on two frictionless chords OA and OB of a vertical circle, at the same instant from point O. The ratio of the velocities of 1 and 2 respectively, when they reach the circumference will be (OB is the diameter)

- sinα
- tanα
- cosα
- none of these

**Q.**A ball is thrown vertically downwards with speed 30 m/s from the top of the tower of height 80 m. The speed of the ball with which it hits the grounds is (g=10 m/s2)

- 40 m/s
- 60 m/s
- 50 m/s
- 30 m/s

**Q.**A point moving with constant acceleration from A to B in the straight line AB has velocities u and v at A and B respectively. Find its velocity at C, the mid point of AB.

- √v2−u22
- √v2+u22
- √v2+u24
- √v2−u24

**Q.**A train is scheduled to run from Delhi to Pune at an average speed of 80 km/h but due to repairs of track , it looses 2 hrs in the first part of the journey. It then accelerates at a rate of 20 km/h2 till the speed reaches 100 km/h. This speed is now maintained till the end of the journey. If the train now reaches station on time, then find the distance from pune to a point when it starts accelerating.

- 840 km
- 640 km
- 550 km
- Data insufficient

**Q.**A ball A is dropped from a 44.1 m high cliff. Two seconds later, another ball B is thrown downward from the same place with some initial speed. The two balls reach the ground together. Find the speed with which the ball B was thrown.

[ Take g=9.8 m/s2 ]

- 39.2 m/s
- 44.1 m/s
- 24.6 m/s
- 12.6 m/s

**Q.**Two particles P and Q initially separated by 400 m, are moving towards each other along a straight line as shown in figure with ap=2 m/s2, up=15 m/s and aQ=4 m/s2, uQ=25 m/s. Calculate the time when they meet.

- 203 s
- 403 s
- 503 s
- 703 s

**Q.**

A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of the tower is

(g=10 m/s2)

- 90 m
- 45 m
- 55 m
- 60 m

**Q.**A particle is dropped from the top of a building . When it crosses a point 5 m below the top , another particles starts to fall from a point 25 m below the top , both particles reach the bottom of the building simultaneously. The height of the building is :

[g=10 m/s2]

- 45 m
- 35 m
- 25 m
- 50 m

**Q.**When a ball is thrown vertically upwards with velocity v0, it reaches a maximum height of h. With what velocity should the ball be thrown to reach half the maximum height.

- √2v0
- √0.5v0
- √1.5v0
- √3v0

**Q.**A ball is thrown vertically upward from the top of a tower. Velocity at depth h from the point of projection is twice of the velocity at height h above the point of projection. Find the maximum height reached by the ball.

- 2h
- 3h
- (53)h
- (43)h

**Q.**If a body travels half its total path in the last second of its fall from rest, find the time of its fall. Take g=10 m/s2.

- 3.4 s
- 4.4 s
- 4.8 s
- 5.4 s

**Q.**A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of 1 m each will then be

- All equal, being equal to √2g second
- In the ratio of the square roots of the integers 1, 2, 3.....
- In the ratio of the difference in the square roots of the integers i.e. √1, (√2−√1), (√3−√2), (√4−√3)...
- In the ratio of the reciprocal of the square roots of the integers i.e., 1√1, 1√2, 1√3, 1√4

**Q.**The driver of a car moving at a speed of 10 m/s applies the brakes after noticing an obstacle. The retardation due to the brakes is 4 m/s2. What is the distance travelled by the car till it stops?

- 12 m
- 12.5 m
- 10 m
- 11 m

**Q.**Two balls are dropped from different heights at different instants. Second ball is dropped 2 s after the first ball. If both the balls reach the ground simultaneously, after 5 s of dropping the first ball, then the difference between the initial heights of the two balls will be

(Take g=9.8 m/s2)

- 58.8 m
- 78.4 m
- 98.0 m
- 117.6 m

**Q.**A body initially moving with a velocity of 4 ms–1, attains a velocity of 20 ms–1 in 4s . The acceleration of the body is:

(Assume constant acceleration)

- 1 ms−2
- 3 ms−2
- 2 ms−2
- 4 ms−2

**Q.**A stone is dropped from a balloon going up with a uniform velocity of 5.0 m/s. If the balloon was 50 m high when the stone was dropped, find its height when stone hits the ground.

- 50 m
- 68.5 m
- 78.5 m
- 40 m

**Q.**Two cars A and B are at rest and at the same point initially. If A starts with uniform velocity of 40 m/s and B starts in the same direction with constant acceleration of 4 m/s2 , then B will catch A after the time

- 10 s
- 20 s
- 30 s
- 35 s

**Q.**Drops of water fall at regular intervals from the roof of a building of height H=16 m. The first drop strikes the ground at the same moment when the fifth drop detaches itself from the roof. The distances between the different drops in air as the first drop reaches the ground are

- 1 m, 4 m, 7 m, 10 m
- 1 m, 3 m, 5 m, 7 m
- 1 m, 4 m, 9 m, 16 m
- None of the above

**Q.**An object is projected upwards with a velocity of 100 m/s. It will strike the ground after

- 10 sec
- 20 sec
- 15 sec
- 5 sec

**Q.**A bike racer starting from rest is riding with constant angular acceleration on a circular track. The ratio of angular displacement of the cyclist during 5th and 7th second is given by

- 7:9
- 4:5
- 9:13
- 8:10

**Q.**A ball thrown vertically upwards with a speed of 19.6 ms−1 from the top of a tower returns to the earth in 6 s. Find the height of the tower.

(Take g=9.8 m/s2)

- 27.8 m
- 58.8 m
- 70 m
- 92 m

**Q.**A car starts form rest and accelerates uniformly over a time of 5 seconds for a distance of 110 m. The acceleration of the car will be

- 3.26 ms−2
- 9.8 ms−2
- 8.8 ms−2
- 7.29 ms−2

**Q.**A frictionless wire is fixed between A and B inside a ring of radius R. A bead slips along the wire. The time taken by the bead to slip from A to B will be

- 2√R/g
gR/√gcosθ

2√gRgcosθ

- 2√gRcosθg

**Q.**A body starts from rest and travels s m in 2nd second, then, acceleration is

- 3s2 m/s2
- 2s m/s2
- 2s3 m/s2
- 3s m/s2

**Q.**A bullet of mass 0.1 kg is fired on a wooden block to pierce through it, but it stops after moving a distance of 50 cm into it. If the velocity of bullet before hitting the wood is 10 m/s and it slows down with uniform deceleration, then the magnitude of effective retarding force on the bullet is x N. The value of x to the nearest integer is

**Q.**A body travels for 15 sec starting from rest with constant acceleration. If it travels S1, S2 and S3 in the first 5 seconds, second 5 seconds and next five seconds respectively. The relation between S1, S2 and S3 is

- S1=S2=S3
- 5S1=3S2=S3
- S1=13S2=15S3
- S1=15S2=13S3

**Q.**A balloon is released from the ground (h=0 m) from rest and starts accelerating with acceleration 5 m/s2 vertically upwards. An object is dropped from the balloon, when it is at height 10 m from the ground. If the balloon is released at t=0 sec, then find the time when the object reaches the ground. Take g=10 m/s2.

- 2.73 s
- 4.73 s
- 4 s
- 2 s

**Q.**A particle travels 10 m in first 5 s and 10 m in next 3 s. Assuming constant acceleration what is the distance travelled in next 2 sec

- 8.3 m
- 9.3 m
- 10.3 m
- None of the above