Q. A stone is dropped from a height $h$. Simultaneously, another stone is thrown up from the ground which reaches a height $4\ h$. The two stones cross other after time
A) $\sqrt{\begin{array}{r}
\frac{h}{8g} \\
\
\end{array}}$
B) $\sqrt{8gh}$
C) $\sqrt{2gh}$
D) $\sqrt{\begin{array}{r}
\frac{h}{2g} \\
\
\end{array}}$
Explanation
(a) For first stone $u = 0$ and
For second stone $\frac{u^{2}}{2g}4h \Rightarrow u^{2} = 8gh$
$\therefore u = \sqrt{8gh}$
Now, $h_{1} = \frac{1}{2}gt^{2}$
$h_{2} = \sqrt{\begin{array}{r}
8ght - \frac{1}{2}gt^{2} \\
\
\end{array}}$
$u = \sqrt{8gh}$
Where,$t =$time cross each other
$\therefore h_{1} + h_{2} = h$
$\Rightarrow \frac{1}{2}gt^{2} + \sqrt{8ght} - \frac{1}{2}gt^{2} = h \Rightarrow t = \frac{h}{\sqrt{8gh}}$=$\sqrt{\begin{array}{r}
\frac{h}{8g} \\
\
\end{array}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.