Question

A body is released from the top of a tower of height h metre. It takes T seconds to reach the ground. Where is the ball at the time T/2 seconds ? (A) at h/4 metre from the ground (B) at h/2 metre from the ground (C) at 3h/4 metre from the ground (D) depends upon the mass of the ballA body is released from the top of a tower of height h metre. It takes T seconds to reach the ground. Where is the ball at the time T/2 seconds ? (A) at h/4 metre from the ground (B) at h/2 metre from the ground (C) at 3h/4 metre from the ground (D) depends upon the mass of the ball

Answer 1

Explanation:

Let the acceleration be 'g'

Then initial velocity = 0 m/s [in T seconds]

Let the distance be 'h' m.

We know,

Equation of motion:

s = ut + 1/2 gT²

Then,

h = 1/2gT² ......(1)

h1 = 1/2gT²/9'......(2)

Substitute (1) and (2):

We'll get h1 = h9

∴ The distance from ground is h-h/9 = 8h/9 ←(Answer)