Question

A stone is released from the top of a tower 125 m high. Then find:

(1) the time with which the stone strikes the ground.

(2) the final velocity of the stone. (Take g = 10 m/s²)

Answer 1

We are given that: Height of tower (h) = 125 m

Initial velocity (u) = 0

g = 10 m/s²

(1) From 2nd equation of motion

For downward motion: h = ut + \(\frac{1}{2}\)gt²

⇒ 125 = 0 x t + \(\frac{1}{2}\) x 10 x t² = 125 = 5t² 

⇒ t² = 25 

⇒ t = 5sec²

(2) From 1st equation of motion,

v = u + gt

⇒ u = 0 + 10 x 5

⇒ v = 50 m/s