Question

A particle is projected with velocity $2\sqrt{gh}$ 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. $\frac{2h}{g}$
• B. $\sqrt{\frac{2h}{g}}$
• C. $\sqrt{\frac{h}{g}}$
• D. $2\sqrt{\frac{h}{g}}$

Answer 1

Answer: (D)

Let $t$ be the time interval. Then,

$2h=\left(u_x\right)(t)$
or $u_x=\frac{2h}{t}$ ...(i)
Further, $h=u_{y}t-\frac{1}{2}g{t}^2$
or $g{t}^2-2u_{y}t+2h=0$
$t_1=\frac{2u_y+\sqrt{4u_y^2-8gh}}{2g}$
and $t_2=\frac{2u_y-\sqrt{4u_y^2-8gh}}{2g}$
$t=t_1-t_2=\frac{\sqrt{4u_y^2-8gh}}{g}$
or $u_y^2=\frac{g^2{\left(t\right)}^2}{4}+2gh$ ...(ii)
Given, $u_x^2+u_y^2={\left(2\sqrt{gh}\right)}^2$
$\frac{4h^2}{{\left(t\right)}^2}+\frac{g^2{\left(t\right)}^2}{4}+2gh=4gh$
$\frac{g^2}{4}{\left(t\right)}^4-2gh{\left(t\right)}^2+4h^2=0$
${\left(t\right)}^2=\frac{2gh\pm \sqrt{4g^2{h}^2-4g^2{h}^2}}{g^2/2}=\frac{4h}{g}$
or $t=2\sqrt{\frac{h}{g}}$