Question

A cannon ball has a range R on a horizontal plane, such that the corresponding possible maximum heights reached are $H_{1}$ and $H_{2}$. Then, the correct expression for R is

• A. $\frac{\left(H_{1}+H_{2}\right)}{2}$
• B. $\left(H_{1} H_{2}\right)^{1 / 2}$
• C. $2\left(H_{1} H_{2}\right)^{1 / 2}$
• D. $4\left(H_{1} H_{2}\right)^{1 / 2}$

Answer 1

Answer: (D)

The cannon ball will have same horizontal range for angle of projection $\theta$ and $\left(90^{\circ}-\theta\right)$. So
$ H_{1}=\frac{u^{2} \sin ^{2} \theta}{2 g}$ and
$H_{2}=\frac{u^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 g}$
$=\frac{u^{2} \cos ^{2} \theta}{2 g}$
$\therefore H_{1} H_{2}=\frac{1}{4}\left(\frac{u^{2} \sin \theta \cos \theta}{g}\right)^{2}$
$=\frac{1}{4} \times \frac{R^{2}}{4}\left(\because R=\frac{u^{2} \sin 2 \theta}{g}\right)$
or $ R=4 \sqrt{H_{1} H_{2}}$