Question

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

Answer 1

Answer:

Initial velocity of stone= Velocity of balloon at height 'h'

Now for balloon,

v^2 - u^2 = 2as

or, v^2 = 2(g/8)h

or, v = ((gh)^(0.5))/2

For stone,

s = ut + a(t^2)/2 (u = v)

or, -h = (((gh)^(0.5))/2)t - g(t^2)/2

(u -> UP

s -> DOWN

a -> DOWN)

or, g(t^2)/2 - (((gh)^(0.5))/2)t - h = 0

or, t

= ((((gh)^(0.5))/2)+ (gh/4 + 2gh)^(0.5))/g (Using quadratic formula and taking + sign)

= ((((gh)^(0.5))/2) + (9gh/4)^(0.5))/g

= ((((gh)^(0.5))/2 + (1.5)×((gh)^(0.5)))/g

= 2×((gh)^(0.5))/g

= 2×((h/g)^(0.5)) ( ANSWER )

I know it is a bit confusing and complicated but please try to understand.

PLEASE MARK IT AS BRAINLIEST ANSWER!!