Question

A balloon rises from rest with a constant acceleration g/8.A stone is released from it when it has risen to a height h.The time taken by the stone to reach the ground is?

Answer 1

Velocity after rising height h
v = sqrt(2aS) ………[∵ initial velocity of balloon is zero]
v = sqrt(2 × g/8 × h)
v = sqrt(gh/4)
This is the initial velocity of stone.

Time taken by stone rise extra height and come to it’s point of projection is
t1 = 2u / g
= 2 × sqrt(gh/4) / g
= sqrt(h / g)

Final velocity of stone when it touches the ground
v’ = sqrt(u^2 + 2aS)
= sqrt[(gh/4) + 2gh]
= sqrt[2.25 gh]
= 1.5 × sqrt(gh)

Time taken by stone to fall down from point of projection
t2 = (v’ - u) / a
= [(1.5 × sqrt(gh)) - sqrt(gh/4)] / g
= sqrt(gh) [1.5 - 0.5] / g
= sqrt(h / g)

Total time taken for stone to reach the ground
T = t1 + t2
= 2 sqrt(h / g)