Question

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.the value of v which would enable the stone B to meet the stone A midway (at mid point
between their initial positions is:
(A) 2 gh
(B) 2√gh
(C) √gh
(D) √2gh​

Answer 1

Answer:

v = √gh

Explanation:

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 say  t time they meet

S = ut + (1/2)at²

a = g ( acceleration due to gravity)

Distance covered by stone A = 0 + (1/2)gt²

Distance covered = h/2  ( mid point)

=> h/2 = 1/2gt²

=> h = gt²

=> t² = h/g

=> t = √h/g

Stone B initial velocity = v

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

h/2 = vt  - (1/2)gt²

putting   t = √h/g

=> h/2 = v√h/g - (1/2)gh/g

=> h = v√h/g

=> v = h√g/√h

=> v = √gh

option C is correct

Answer 2

Explanation:

answer is under root hg