 # A stone is dropped from a height h.simultaneously another stone is thrown up from the ground which reaches the height 4h.the two stones cross each other after a time-------------

10 years ago

For upper stone which is dropped downwards the distance traveled is

S1 = ut + 1/2  g t^2

Here u = 0

g = accln due to gravity = 9.8 m/s2

So S1 = 1/2g t2

Distance of stone from ground is = h - S1 = h -  0.5 g t2    ....(1)

Now for stone thrown upwards such that it goes to height 4h from ground

Potential energy at top = Kinetic energy from the point of projection

or mg(4h) = 1/2 m u2

or u = (8hg)1/2

Thus S2 = ut  -  0.5 g t2

or S2 =  (8hg)1/2 t - 0.5 g t2   ........(2)

Now for the two stones to meet

h - S1 = S2

or h -  0.5 g t2   =  (8hg)1/2 t - 0.5 g t2                (from 1 and 2)

or h = (8hg)1/2 t

or t = ( h/8g) 1/2

I thnk this could help u...

With Regards,

RONIT MOHANTY® 10 years ago

initial vel of stone thrown up from ground v(say)

0²-v²=-2g(4h)

=>v=√(8gh)

since both are droped symultameously

hence relative accleration is 0

so vel. of thrown up will seem constant to vel to the dropped one

hence time=dist/speed=h/√(8gh)=√(h/8g) ans. Kushagra Madhukar
askIITians Faculty 629 Points
2 years ago
Dear student,
Please find the attached solution to your problem
Let the initial velocity of stone which is thrown from the ground u2.
At maximum height 4h, Velocity v2=0
Using v2 = u2 2as
0 = u222g×4h
u2= √8gh
Let h1 be the distance covered by the stone which is dropped from height h,
Let h2 be the distance covered by the stone which is thrown up from the ground.
Let at time t both stone crossed each other.
Using s = ut + 1/2at2,
h1 = 1/2gt2
h1= 1/2gt2 ....1
h2= u2t − 1/2gt2 ....2
Adding h1 and h2 gives the height h.
Adding equations 1 and 2,
h = 1/2gt2 + u2t − 1/2gt2
h = u2t
Using the value ofu2= √8gh
h =√8gh * t
t =√h/8g
Hope it helps.
Thanks and regards,
Kushagra