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

solution:

Range, R = 3 km

Angle of projection, $\theta=30^{\circ}$

Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$

Horizontal range for the projection velocity $u_{0}$, is given by the relation:

$R=\frac{u_{0}^{2} \sin 2 \theta}{\mathrm{g}}$

$3=\frac{u_{0}^{2}}{\mathrm{~g}} \sin 60^{\circ}$

$\frac{u_{0}{ }^{2}}{\mathrm{~g}}=2 \sqrt{3}$ $\ldots(i)$

The maximum range $\left(R_{\max }\right)$ is achieved by the bullet when it is fired at an angle of $45^{\circ}$ with the horizontal, that is,

$R_{\max }=\frac{u_{0}^{2}}{g}$ $\ldots(i i)$

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

$R_{\max }=3 \sqrt{3}=2 \times 1.732=3.46 \mathrm{~km}$

Hence, the bullet will not hit a target $5 \mathrm{~km}$ away.
Editor