Question

A body is released from top of a tower of height h and takes t sec to reach ground. The position of body at t/2 sec

Answer 1

Answer:

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)

Good Studies!