Question

A particle is projected with velocity $2\sqrt{g h}$ so that it just clears two walls of equal height $h$ , which are at a distance of $2h$ from each other. What is the time interval of passing between the two walls?

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

Answer 1

Answer: (D)

Let $t$ be the time interval. Then,

$2h=\left(u_{x}\right)\left(\right.t\left.\right)$
or $u_{x}=\frac{2 h}{t}$ ...(i)
Further, $h=u_{y}t-\frac{1}{2}gt^{2}$
or $gt^{2}-2u_{y}t+2h=0$
$ \, t_{1}=\frac{2 u_{y} + \sqrt{4 u_{y}^{2} - 8 g h}}{2 g}$
and $t_{2}=\frac{2 u_{y} - \sqrt{4 u_{y}^{2} - 8 g h}}{2 g}$
$t=t_{1}-t_{2}=\frac{\sqrt{4 u_{y}^{2} - 8 g h}}{g}$
or $u_{y}^{2}=\frac{g^{2} \left(t\right)^{2}}{4}+2gh$ ...(ii)
Given, $u_{x}^{2}+u_{y}^{2}=\left(2 \sqrt{g h}\right)^{2}$
$ \, \frac{4 h^{2}}{\left(t\right)^{2}}+\frac{g^{2} \left(t\right)^{2}}{4}+2gh=4gh$
$\frac{g^{2}}{4}\left(t\right)^{4}-2gh\left(t\right)^{2}+4h^{2}=0$
$\left(t\right)^{2}=\frac{2 g h \pm \sqrt{4 g^{2} h^{2} - 4 g^{2} h^{2}}}{g^{2} / 2}=\frac{4 h}{g}$
or $t=2\sqrt{\frac{h}{g}}$