Question

A balloon rises from rest with constant acceleration g/g , a stone is dropped when the balloon reaches a height of 39.2 m. Find the time taken by the stone to reach the Ground.

Answer 1

u=0m/s
s=39.2m
a=9.8m/s
v=?
V2-U2=2as
V2-02=2×9.8×39.2
V2=768.32
V=28m/s

V=U+at
28 =0+9.8t
t=28/9.8
t=2.857s
Hope it helps you ☺☺

Answer 2

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)