Question

A stone A is dropped 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 midpoint) between their initial position is:

• A. 2 gh
• B. $2\sqrt{gh}$
• C. $\sqrt{gh}$
• D. $\sqrt{2gh}$

Answer 1

The correct option is C $\sqrt{gh}$

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 + $\frac{1}{2}$ at²

a = g ( acceleration due to gravity)

Distance covered by stone A = 0 + $\frac{1}{2}$ gt²

Distance covered = $\frac{h}{2}$ ( mid point)

=> $\frac{h}{2}$ = $\frac{1}{2}$gt²

=> h = gt²

=> t² = $\frac{h}{g}$

=> t = $\sqrt{\frac{h}{g}}$

Stone B initial velocity = v

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

$\frac{h}{2}$ = vt - $\frac{1}{2}$gt²

putting t = $\sqrt{\frac{h}{g}}$

=> $\frac{h}{2}$ = v$\sqrt{\frac{h}{g}}$ - $\frac{1}{2}$$\times$$\frac{gh}{g}$

=> h = $v\sqrt{\frac{h}{g}}$

=> v = $h\frac{\sqrt{g}}{\sqrt{h}}$

=> v = $\sqrt{gh}$