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: (C)

The value of v is √gh.

Answer 2

Let the origin be at the ground and upward direction be positive.

For stone A (dropped from rest at height h): Initial position $y_{A0} = h$, initial velocity $u_A = 0$, acceleration $a_A = -g$. Position at time t: $y_A(t) = h - \frac{1}{2}gt^2$.

For stone B (thrown up from ground with velocity v): Initial position $y_{B0} = 0$, initial velocity $u_B = v$, acceleration $a_B = -g$. Position at time t: $y_B(t) = vt - \frac{1}{2}gt^2$.

They meet at the midpoint, which is at height $h/2$. Let the time of meeting be T. So, $y_A(T) = h/2$ and $y_B(T) = h/2$.

From stone A's motion: $\frac{h}{2} = h - \frac{1}{2}gT^2$ $\frac{1}{2}gT^2 = h - \frac{h}{2}$ $\frac{1}{2}gT^2 = \frac{h}{2}$ $gT^2 = h \implies T = \sqrt{\frac{h}{g}}$.

From stone B's motion: $\frac{h}{2} = vT - \frac{1}{2}gT^2$ Substitute $\frac{1}{2}gT^2 = \frac{h}{2}$: $\frac{h}{2} = vT - \frac{h}{2}$ $h = vT$ $v = \frac{h}{T}$

Substitute the value of T: $v = \frac{h}{\sqrt{\frac{h}{g}}} = h \sqrt{\frac{g}{h}} = \sqrt{\frac{h^2 g}{h}} = \sqrt{gh}$.