Question

A bullet fired at an angle of ${30}^o$ with the horizontal hits the ground $3.0km$ away. By adjusting its angle of projection, can one hope to hit a target $5.0km$ away? Assume the muzzle speed to be fixed, and neglect air resistance.

Answer 1

Range, $R=3km$

Angle of projection, $={30}^o$

Acceleration due to gravity, $g=9.8m/s^2$
Horizontal range for the projection velocity $u_0$, is given by the relation:
$R=\frac{u_o^{2}Sin2\theta }{g}$
$3=\frac{u_o^{2}Sin{60}^0}{g}$

$\frac{u_o^2}{g}=2\sqrt{3}$ .......(i)

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:
$R_{max}=2\times 1.732=3.46$ km
Hence, the bullet will not hit a target $5$ km away.