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)

$H _{1}=\frac{ u ^{2} \sin ^{2} \theta}{2 g }$
and $H _{2}=\frac{ u ^{2} \sin ^{2}(\pi / 2-\theta)}{2 \sigma}=\frac{ u ^{2} \cos ^{2} \theta}{2 g }$
$H _{1} H _{2}=\frac{ u ^{2} \sin ^{2} \theta}{2 g } \times \frac{ u ^{2} \cos ^{2} \theta}{2 g }$
$=\frac{\left( u ^{2} 2 \sin \theta \cos \theta\right)^{2}}{16 g ^{2}}=\frac{ R ^{2}}{16}$
$\therefore R =4 \sqrt{ H _{1} H _{2}}$