Question

A ball is released from top of tower of a height h. It takes time T to reach on the ground. What is the position of ball in time T/2?

Answer 1

Answer:

S = ut + 1/2 at²

 h = o (t) +1/2×10×t²

 h = 5 t²

when time is T/2

s = (0)t + 1/2 a(t/2)²

s = 1/2×10× t²/4

s = 5 t²/4

here h = 5 t² hence

 S =  h/4

Explanation:

Answer 2

Initial velocity of the ball = 0

acceleration = g

The distance traveled in time T is given as h

Applying the equations of motion

s= ut+1/2gt² we get;

s= 1/2 gt²         [intial velocity u=0 in this case]

so we can say that the distance traveled is directly proportional to the square of the time taken to travel that distance

ie; s∝ t²

⇒ H∝ T² ....... (1)

if h is the distance travelled in time T/2 then

 h∝ (T/2)²...... (2)

taking the ratio of (1) and (2) we get

h/ H = T²/4 ÷ T²

h/H = 1/4

⇒ h= H/4