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, what is the farthest distance can one hope to hit.

• A. $3.46\, km$
• B. $3.82 \,km$
• C. $4.12 \,km$
• D. $4.46 \,km$

Answer 1

Answer: (A)

$R=\frac{u^{2} \sin 2 \theta}{g}$
Here $\theta=30^{\circ}, R=3 \,km ,$ so $\frac{u^{2}}{g}=\frac{3}{\sin 60^{\circ}} \,km$
But $R_{\max }=\frac{u^{2}}{g}=\frac{3 \times 2}{\sqrt{3}}=2 \sqrt{3}=3.46 \,km$