Question

A balloon rises from rest on the ground with constant accelaration g/3.A stone is dropped when the balloon has rises to aheight 60m.the time taken by the stone to reach the ground is..?

Answer 1

we have
v2 - u2 = 2as = 2(g/8).h
here, u = 0 (for balloon) so,
the velocity of the balloon at the height h is 
v = [2(g/8).h]1/2 = (1/2).[(g/8).h]1/2 

now, initial velocity of the stone of the stone at height h will be
u' = v = [(gh)/2]1/2 and will be in upward direction
so, height will be
h = - u't + (1/2)gt2 

so, by using value of u' we get
h = -[(gh)/2]1/2.t + (1/2)gt2 
or
[multilying by 2 and rearranging] 
gt2 - (gh)1/2.t - 2h = 0 

The above us a quadratic equation [ax2 + bx + c = 0] which cane be solved, givign usthe value of time 't' as
[ x = {-b ∓ (b2 - 4ac)1/2 } / 2a ]

t = 2 (h/g)1/2

Answer 2

Answer:

Step-by-step 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)