Question

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

Answer 1

Step 1: Find muzzle initial speed.
Given $R=3\text{km}$
$\theta ={30}^{\circ }$
Horizontal range for projection velocity 𝑢 is given by,
$R=\frac{u^2\sin 2\theta }{g}$
$3=\frac{u^2\sin {60}^{\circ }}{g}$
$\frac{u^2}{g}=2\sqrt{3}$
$u^2=\left(2\sqrt{3}g\right)....\left(1\right)$

Step 2: Find Maximum horizontal range for same projectile.
Maximum range can be achieved by the bullet when it is fired at an angle of ${45}^{\circ }$ with the horizontal,
$R_{max}=\frac{u^2}{g}$
$R_{max}=2\sqrt{3}=3.46\text{km}$

Final Answer: As maximum range $R_{max}<5\text{km}$, So the bullet cannot hit a target 5 km away.