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 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 that, $H_1=\frac{u^2{\sin }^2\theta }{2g}$
and $H_2=\frac{u^2{\sin }^2(90-\theta )}{2g}$
$\Rightarrow H_2=\frac{u^2{\cos }^2\theta }{2g}$
$\because H_1{H}_2=\frac{u^2{\sin }^2\theta }{2g}\times \frac{u^2{\cos }^2\theta }{2g}$
$=\frac{4}{4}\times \frac{U^2{\sin }^2\theta }{2g}\times \frac{U^2{\cos }^2\theta }{2g}$
$=\frac{{\left(U^2\sin 2\theta \right)}^2}{16g^2}$ $[\because {\sin }^2\theta =2\sin \theta \cos \theta ]$
$=\frac{R^2}{16}$
$R^2=6H_1{H}_2$
$\therefore R=4\sqrt{H_1{H}_2}$