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 $(\frac{\pi }{2}-\theta )$ with the horizontal, it reaches the 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. $R=4(H_1-H_2)$
• C. $R=4(H_1+H_2)$
• D. $R=\frac{H_1^2}{H_2^2}$

Answer 1

The correct option is A $R=4\sqrt{H_1{H}_2}$

We know
$H_1=\frac{u^2{\sin }^2\theta }{2g}$
$H_2=\frac{u^2{\sin }^2(90-\theta )}{2g}=\frac{u^2{\cos }^2\theta }{2g}$
$H_1{H}_2=\frac{u^2{\sin }^2\theta }{2g}\times \frac{u^2{\cos }^2\theta }{2g}=\frac{(u^2\sin 2\theta {)}^2}{16g^2}=\frac{R^2}{16}$
$\therefore R=4\sqrt{H_1{H}_2}$