Question

A bullet fired at an angle of 30° 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 the fixed, and neglect
air resistance.

Answer 1

Explanation:

HEY mate.......

$range \: = 3km \: \\ \\ angle \: of \: projection \: = 30 \: degree \\ \\ acceleration \: due \: to \: gravity = 9.8m | {s}^{2} | \\ \\ horizontal \: range \: of \: projection \: velocity \: u0 \: \\ \\ is \: given \: by \: the \: relation \: \\ \\ r = \frac{ {u0}^{2} \sin(2thetha) }{g} \\ \\ 3 = \frac{ {u0}^{2} \sin(60degree) }{g} \\ \\ \frac{ {u0}^{2} }{g} = \sqrt[2]{3} ......(1eqn)$

The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of 45 with the horizontal,

that is, Rmax = u o^2 / g ....(ii)

On comparing equations (i) and (ii), we get:

R mma =2×1.732=3.46 km

Hence, the bullet will not hit a target 5 km away.

HOPE IT HELPS YOU