Question

A ball is released from the top of a tower of height ‘h’ meters. It takes ‘t’ seconds to reach the ground. Where is the ball at t/2 seconds?
a) At h/2 meters from the ground
b) At h/4 meters from the ground
c) At 3h/4 meters from the ground
d) Depends on the mass and volume of the ball

Answer 1

Answer:

Explanation:

Since the ball is released from top of a tower H metre high, its initial velocity u = 0 m/s. It takes ‘T' seconds to reach the ground. Height H and T are related through the relation: s = u t +½ g t². Taking the top of the tower as the origin and downward direction as positive, we have,

H = 0 × T+ ½ g T² = = > H = ½ g T² –––––––––––(1)

The position of the ball at time T/2 after its release from top of the tower be H', then

H' = 0× T + ½ g (T/2)² = ½ g T²/4 = H/4 –––––––––(2)

So the ball would have travelled only ¼th of the height H of the tower from the origin ie the top of the tower and would still be (H- H/4=) ¾ H above the ground at time T/2.

Answer 2

Explanation:

answer will be option c