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, heights $H_{1}$ 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)

$H_{1}=\frac{u^{2} \sin ^{2} \theta}{2 g}$
and $H_{2}=\frac{u^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 g}=\frac{u^{2} \cos ^{2} \theta}{2 g}$
$H_{1} H_{2}=\frac{u^{2} \sin ^{2} \theta}{2 g} \times \frac{u^{2} \cos ^{2} \theta n}{2 g}$
$=\frac{\left(u^{2} \sin 2 \theta\right)^{2}}{16 g^{2}}=\frac{R^{2}}{16}$
$\therefore R=4 \sqrt{H_{1} H_{2}}$