Question

A body is dropped from a height h and a second body is thrown vertically up from
the ground with a velocity of gh simultaneously. At what height from the ground do
they meet each other?

Answer 1

Answer:

A stone A is dropped from rest from a height h above the ground. A second stone B is simultaneously thrown vertically up from a point on the ground with velocity v. The line of motion of both the stones is same.

Stone A  is dropped

so initial velocity  u = 0

Let's say  t time they meet

S = ut + 21 at²

a = g ( acceleration due to gravity)

Distance covered by stone A = 0 + 21 gt²

Distance covered =  2h ( mid point)

=> 2h = 21gt²

=> h = gt²

=> t² = gh

=> t = gh

Stone B initial velocity = v

a = -g ( as it is going against  gravity)

2h = vt  - 21gt²

putting   t = gh

=> 2h = vgh - 21×ggh

=> h = vgh

=> v = hh

Answer 2

Answer:

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