Question

16.
A balloon rises from rest on the ground with constant acceleration 2. A stone is dropped when the
balloon has rises to a height 60 metre. The time taken by the stone to reach the ground is.
Motion with variable acceleration & calculus​

Answer 1

Answer:

Explanation:

Answer:

Explanation:

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)