Question

A particle is projected with velocity $2\sqrt{gh}$ so that it just clears two walls of equal height $h$ which are distance $2h$ from each other. The time interval for which the particle travels between the two walls is

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

Answer 1

The correct option is A $2\sqrt{\frac{h}{g}}$

$h=usin\theta \times t-\frac{1}{2}g{t}^2$
$\frac{1}{2}g{t}^2-u\sin \theta \times t+h=0$
$\Rightarrow t_1+t_2=\frac{2u\sin \theta }{g}\Rightarrow t_1\cdot t_2=\frac{2h}{g}$

$\Rightarrow u\cos \theta \Delta t=2h$
$\Rightarrow \Delta t=\frac{2h}{u\cos \theta }\Rightarrow \Delta t=\frac{2h}{2\sqrt{gh}\cos \theta }$ ....... (1)
$\Rightarrow t_2-t_1=\sqrt{(t_1+t_2{)}^2-4t_1{t}_2}$
$\Rightarrow \frac{4h^2}{u^2{\cos }^2\theta }=\frac{4u^{2}si{n}^2\theta }{g^2}-\frac{8h}{g}$
put $u=2\sqrt{gh}$

$\Rightarrow \frac{h^2}{gh{\cos }^2\theta }=\frac{16ghsi{n}^2\theta }{g^2}-\frac{8h}{g}$
$\Rightarrow \frac{1}{{\cos }^2\theta }=16{\sin }^2\theta -8$
$\Rightarrow 1=16{\sin }^2\theta {\cos }^2\theta -8{\cos }^2\theta$

$\Rightarrow$ putting ${\cos }^2\theta =x$
$1=16(1-x)x-8x$
$\Rightarrow 1=16x-16x^2-8x$
$\Rightarrow 16x^2-8x+1=0$
$\Rightarrow 16x^2-4x-4x+1=0$
$\Rightarrow 4x(4x-1)-1(4x-1)=0$
$\Rightarrow (4x-1{)}^2=0$
$\Rightarrow x=\frac{1}{4}$
$\Rightarrow co{s}^2\theta =\frac{1}{4}$
$\Rightarrow cos\theta =\pm \frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{3}$ & $\frac{2\pi }{3}$

from equation 1
$\Delta t=\frac{2}{2\sqrt{gh}cos\theta }$
$=2\sqrt{\frac{h}{g}}or-2\sqrt{\frac{h}{g}}$
So, at $\theta =\frac{\pi }{3}$
$\Delta t=2\sqrt{\frac{h}{g}}$