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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-------------

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-------------

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3 Answers

Ronit Mohanty
33 Points
12 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®Smile

pankaj kumar raman
51 Points
12 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 628 Points
3 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 = u22​ − 2g×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 + 1/2​at2,
⇒ −h1 ​= −1/2​gt2
⇒h1​= 1/2​gt2 ....1
⇒h2 ​= u2​t − 1/2​gt2 ....2
Adding h1​ and h2​ gives the height h.
∴ Adding equations 1 and 2,
⇒h = 1/2​gt2 + u2​t − 1/2​gt2
⇒h = u2​t
Using the value ofu2​ = √8gh
⇒h =√8gh * t
⇒t =√h/8g
Hope it helps.
Thanks and regards,
Kushagra​​

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