Question

In a projectile motion, given $H=\frac{R}{2}=20m.$ Here, H is maximum height and R the horizontal range. For the given condition match the following two columns.

Answer 1

$\frac{H}{R}=\frac{\frac{u^2{\sin }^2\theta }{2g}}{\frac{u^2\sin 2\theta }{g}}=\frac{\tan \theta }{4}$
Given, $\frac{H}{R}=\frac{1}{2}$
$\Rightarrow \frac{1}{2}=\frac{\tan \theta }{4}$
$\tan \theta =2$
$\Rightarrow$ Ratio of vertical component of velocity and horizontal component of velocity is 2 as $\frac{u_y}{u_x}=\tan \theta$
Time of flight: $T=\frac{2u\sin \theta }{g}=2\frac{\sqrt{u^2{\sin }^2\theta }}{g}=\frac{2\sqrt{2gH}}{g}=\frac{40}{10}=4$
$\frac{u\sin \theta }{u\cos \theta }=2$
$\Rightarrow \frac{20}{u\cos \theta }=2$
$\Rightarrow u\cos \theta =10$