Question

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. $\sqrt{\frac{h}{2g}}$
• B. $\sqrt{\frac{h}{8g}}$
• C. $\sqrt{8hg}$
• D. $\sqrt{2hg}$

Answer 1

The correct option is B $\sqrt{\frac{h}{8g}}$

For B: $u$ is such that stone B reaches uh.
0 at $ma{x}^m$ pt

Hence, $v^2=u^2-2g_B$
$u^2=2g\left(uh\right)\Rightarrow u=\sqrt{8gh}.....\left(1\right)$

Now $e{q}^n$ for B: $S_B=ut-\frac{1}{2}g{t}^2.....\left(2\right)$

$e{q}^n$ for A: $S_A=\frac{1}{2}g{t}^2.....\left(3\right)$
$S_A+S_B=h=ut\Rightarrow t=\frac{h}{u}$
From (1), $u=\sqrt{8gh}\Rightarrow t=\sqrt{\frac{h}{8g}}$