Question

A bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3 \,km$ away. By adjusting its angle of projection, can one hope to hit a target $5 \,km$ away. Assume the muzzle speed to be same and the air resistance is negligible

• A. possible to hit a target 5 km away
• B. not possible to hit a target 5 km away
• C. prediction is not possible
• D. none of the above

Answer 1

Answer: (B)

It is not possible to hit beyond maximum range. The body covers a horizontal distance $A B$ during its flight. This horizontal range is given by

$R=\frac{u^{2} \sin 2 \theta}{g}\,\,\,...(i)$
where, $u$ is velocity of projection, $\theta$ is angle of projection and $g$ is acceleration due to gravity. For maximum horizontal range $\sin 2 \theta=1$
$\therefore R_{\max }=\frac{u^{2}}{g}\,\,\,...(ii)$
Given, $R=3\, km , \theta=30^{\circ}$
$\therefore $ From Eq. (i)
$\frac{u^{2}}{g}=\frac{R}{\sin 2 \theta}=\frac{3}{\sin 60^{\circ}}=\frac{3 \times 2}{\sqrt{3}}=\sqrt{3} \times 2$
$\therefore \frac{u^{2}}{g}=3.464 \,m$
Hence, maximum range with velocity of projection $u$ cannot be more than $3.464 \,m$. Hence, it is not possible to hit a target $5\, km$ away.