Q. 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) $R = 4(H_{1} - H_{2})$
C) $R = 4(H_{1} + H_{2})$
D) $R = \frac{H_{1}^{2}}{H_{2}^{2}}$
Explanation
(a) $H_{1} = \frac{u^{2}\sin^{2}\theta}{2g}$ and $H_{2} = \frac{u^{2}\sin^{2}{(90 - \theta)}}{2g} = \frac{u^{2}\cos^{2}\theta}{2g}$
$H_{1}H_{2} = \frac{u^{2}\sin^{2}\theta}{2g} \times \frac{u^{2}\cos^{2}\theta}{2g} = \frac{\left( u^{2}\sin{2\theta} \right)^{2}}{16g^{2}} = \frac{R^{2}}{16}$
$\therefore R = 4\sqrt{H_{1}H_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.