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.show that the time of passing between the walls is $2\sqrt{\frac{h}{g}}.$

Answer 1

$h=2\sqrt{gh}\sin \theta t-\frac{1}{2}g{t}^2$

$t^2-4\sqrt{\frac{h}{g}}\sin \theta t+\frac{2h}{g}=0$

$t=\frac{16h}{g}{\sin }^2\theta -\frac{8h}{g}=\frac{8h}{g}\left(2{\sin }^2\theta -1\right)$

$t_2-t_1=\sqrt{\frac{8h\cos \left(180-2\theta \right)}{g}}$

Projectile travelled $2h$ distance is $t_2-t_1$ horizontally –

$2\sqrt{gh}\cos \theta \times \sqrt{\frac{8h\cos \left(180-2\theta \right)}{g}}=2h$

$\cos \theta \sqrt{\frac{8j\cos \left(180-2\theta \right)}{g}}=2h$

$\cos \theta =\sqrt{8\cos \left(180-2\theta \right)}=1$

$-8\left[\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right]=1+{\tan }^2\theta$

${\tan }^4\theta -6{\tan }^2\theta +9=0$

$\tan \theta =\sqrt{3}$

$\theta =60$

$t=\sqrt{\frac{8h}{g}\cos 60}=\sqrt[2]{2\frac{h}{g}}$