Question

A stone projected with a velocity $u$ at an angle $\theta$ with the horizontal reaches maximum height $H_1.$ When it is projected with velocity $u$ at an angle $\left(\frac{\pi }{2}-\theta \right)$ with the horizontal, it reaches maximum height $H_2.$ The relation between the horizontal range $R$ of the projectile, $H_1$ and $H_2$ is :-

• A. $R=4\sqrt{H_1{H}_2}$
• B. $R=4\left(H_1-H_2\right)$
• C. $R=4\left(H_1+H_2\right)$
• D. $R=\frac{H_1^2}{H_2^2}$

Answer 1

Answer: (A)

$H_1=\frac{u^2{\sin }^2\theta }{2g}$
and $H_2=\frac{u^2{\sin }^2(\pi /2-\theta )}{2\sigma }=\frac{u^2{\cos }^2\theta }{2g}$
$H_1{H}_2=\frac{u^2{\sin }^2\theta }{2g}\times \frac{u^2{\cos }^2\theta }{2g}$
$=\frac{{\left(u^{2}2\sin \theta \cos \theta \right)}^2}{16g^2}=\frac{R^2}{16}$
$\therefore R=4\sqrt{H_1{H}_2}$