# Applications of equations of motion

## Trending Questions

**Q.**A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is :

(g=10 m/s2)

- 320 m
- 300 m
- 360 m
- 340 m

**Q.**

A rubber ball is released from a height of $5m$ above the floor. It bounces back repeatedly, always rising to $\frac{81}{100}$ of the height through which it falls. Find the average speed of the ball. (Take $g=10m{s}^{-2}$)

$2.5m{s}^{-1}$

$3.5m{s}^{-1}$

$3.0m{s}^{-1}$

$2.0m{s}^{-1}$

**Q.**An engine of a train, moving with uniform acceleration, passes the signal post with velocity u and the last compartment passes the signal post with velocity v. The velocity with which the middle point of the train passes the signal post is :

- √v2−u22
- √v−u2
- √v2+u22
- √v+u2

**Q.**In a car race on straight road, car A takes time t less than car B at the finish and passes finishing point with a speed v more than that of car B. Both the cars start from rest and travel with constant acceleration α and β respectively. Then v is equal

- (√2αβ)t

- (√αβ)t

- (α+β2)t
- (2α+βα+β)t

**Q.**A particle starts from a point with a velocity of +6 m/s and moves with an acceleration of −2 m/s2. After what time will the particle be at the starting point?

- 2 s
- 6 s
- 4 s
- 5 s

**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 car and a truck move in the same straight line at the same instant of time from the same point. The car moves with a constant velocity of 40 m/s and truck starts with a constant acceleration of 4 m/s2. Find the time t that elapses before the truck catches with the car and also find the greatest distance Smax between them prior to it

- t=5 sec, Smax=100 m
- t=20 sec, Smax=200 m
- t=15 sec, Smax=50 m
- t=8 sec, Smax=100 m

**Q.**A ball is thrown vertically upwards with a velocity u from the ground. The ball attains a maximum height Hmax.. Then find out the displacement at which ball has half of the maximum speed.

- Hmax2
- 32Hmax
- 23Hmax
- 34Hmax

**Q.**

A bullet when first into a target loses half of its velocity after penetrating $20cm$. Further distance of penetration before it comes to rest is

$6.66cm$

$3.33cm$

$12.5cm$

$10cm$

$5cm$

**Q.**A particle starts from a point with a velocity of +6 m/s and moves with an acceleration of −2 m/s2. After what time will the particle be at the starting point?

- 2 s
- 6 s
- 4 s
- 5 s

**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 girl throws a ball vertically upwards with an initial speed of 15 m/s. The ball was released when it was at 2.0 m above the ground. The girl catches it at the same point as the point of projection. What is the maximum height reached by the ball?

- 11.5 m
- 13.5 m
- 15.5 m
- 17.5 m

**Q.**A car accelerates uniformly from 13 ms–1 to 31 ms–1 while entering the motorway, covering a distance of 220 m. Then the acceleration of the car will be:

- 2.9 ms−2
- 1.8 ms−2
- 4 ms−2
- 2.2 ms−2

**Q.**A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is :

(g=10 m/s2)

- 320 m
- 300 m
- 360 m
- 340 m

**Q.**A body starts from rest and moving with an acceleration of 1 ms–2. The displacement of the body in 5 seconds is

- 12.5 m
- 25 m
- 7.5 m
- 15 m

**Q.**An elevator car whose floor to ceiling distance is equal to 2 m starts ascending with constant acceleration 2 m/s2. 5 sec after the start, a bolt begins falling from ceiling of car. Time after which bolt hits floor of elevator starting from fall. (g=10 m/s2)

- 0.7 s
- 1√2 s
- 1.73 s
- 1√3 s

**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.**When a ball is thrown vertically upwards with velocity v0, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity.

- √3v0
- 3v0
- 9v0
- 32v0

**Q.**A ball is thrown vertically upwards with a velocity u from the ground. The ball attains a maximum height Hmax.. Then find out the displacement at which ball has half of the maximum speed.

- Hmax2
- 32Hmax
- 23Hmax
- 34Hmax

**Q.**

A body starts from rest and moves with a uniform acceleration. Find the ratio of the distance covered in the ${\mathrm{n}}^{\text{th}}$ sec to the distance covered in $nsec$.

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

- 45
- 54
- 125
- 512

**Q.**A ball is thrown vertically upwards with velocity of 10 m/s from the top of a tower. It returns to ground after some time with speed of 60 m/s. The height of the tower is

(g=10 m/s2)

- 375 m
- 175 m
- 125 m
- 225 m

**Q.**A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10ms, then maximum height attained by it is (Take g =10 ms2)

- 15 m
- 20 m
- 10 m
- 5 m

**Q.**A stone is dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown, one upwards and other downwards, with the same velocity u and they reach the earth surface after t1 and t2 seconds respectively, then

- t = t1 − t2

- t = t1+t22

- t = √t1t2

- t = t21t22

**Q.**Some parameters are given in Column - I which are to be matched with their correct answers given in Column - II.

Choose the correct option.

Column IColumn IIa)Velocity at t=0 of a particle followingp)αequation x=u(t−8)+α(t−2)2b)Acceleration of a particle moving according to equationq)2αx=4(t−5)+α(t−2)2 at t=0c)Velocity of the particle moving according to equationr)ux=ut+5α t2 ln(1+t2)+α t2 at t=0d)Acceleration of the particle moving according to the equations)u−4αx=ut+5α t2 ln(1+t2)+α t2 at t=0

- a→s b→q c→r d→q
- a→p b→r c→q d→q
- a→p b→r c→s d→q
- a→p b→q c→r d→s

**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.**A particle is thrown vertically upwards. It's velocity at half of the height is 10 m/s, then the maximum height attained by it will be:

- 10 m
- 20 m
- 15 m
- 25 m

**Q.**A particle is dropped under gravity from rest from a height h(g=9.8msec2) and it travels a distance 9h25 in the last second, the height h is

- 100 m
- 122.5 m
- 145 m
- 167.5 m

**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.**From the top of a tall mountain, a stone is dropped. One second later, another stone is projected downwards with velocity 20 ms−1. Find the distance from top where both stones meet.

- 454 m
- 40 m
- 10 m
- 508 m