Question

-RUISES
2. A body is dropped from the top of a tower of
height 100 m. Taking g = 10 m s', calculate the
(a) time taken by the body to reach the ground.
(Ans. 4.47
(b) final velocity with which the body strikes the
ground.
[Ans. 44.7 ms​

Answer 1

Given:-

→Height of the tower = 100m

→Acceleration due to gravity = 10m/s²

To find:-

→Time taken by the body to reach the ground

→Final velocity with which the body strikes the ground.

Solution:-

In this case:-

•Inital velocity of the ball will be zero as it was initially at rest

•Acceleration due to gravity(g) will be 9.8m/s² as the body is freely-falling.

By using the 2nd equation of motion,we get:

=>h = ut+1/2gt²

=>100 = 0(t)+1/2×10×t²

=>100 = 5t²

=>t² = 100/5

=>t² = 20

=>t = √20

=>t = 4.47s

Now,by using the 3rd equation of motion,we get:-

=>v²-u² = 2gh

=>v²-0 = 2(10)(100)

=>v² = 2000

=>v = √2000

=>v = 44.7m/s

Thus:-

→Time taken by the body to reach ground is 4.47 seconds.

→Final velocity of the body with which it strikes the ground is 44.7m/s.