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
$H_1=\frac{u^2{\sin }^2\theta }{2g}$ and
$H_2=\frac{u^2{\sin }^2\left(90^{\circ }-\theta \right)}{2g}$
$=\frac{u^2{\cos }^2\theta }{2g}$
$\therefore H_1{H}_2=\frac{1}{4}{\left(\frac{u^2\sin \theta \cos \theta }{g}\right)}^2$
$=\frac{1}{4}\times \frac{R^2}{4}\left(\because R=\frac{u^2\sin 2\theta }{g}\right)$
or $R=4\sqrt{H_1{H}_2}$