Question

A bullet fired at an angle of $30°$ 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

Step 1: Given data

1. The angle of projection of the bullet is, $v={30}^o$.
2. The horizontal distance traveled by the bullet, $s=3km.$
3. The acceleration due to gravity, $g=9.8m/s^2.$

Step 2: The motion of a projectile

1. When we throw an object (projectile) with an angle to the ground, it travels through a parabolic path and falls down at some distance from the point of the journey.
2. If the projectile is thrown with angle $\theta$ and velocity $u$ with respect to horizontal ground, then the horizontal distance traveled by the projectile is given by $R=\frac{u^2\sin 2\theta }{g}$.
3. The maximum horizontal distance (at $\theta =45°$) traveled by the body is $R_{max}=\frac{u^2}{g}$.

Step 3: Diagram

Step 4: Finding the term $\frac{u^2}{g}$.

Let $u$ be the velocity of the projection of the bullet.

Now,

$$\begin{gathered} R=\frac{u^2\sin 2\theta }{g} \\ or\frac{u^2}{g}=\frac{R}{\sin 2\theta }=\frac{3}{\sin 60°}=\frac{3}{{\displaystyle \frac{\sqrt{3}}{2}}}=3.47 \\ or\frac{u^2}{g}=3.47..................\left(1\right) \end{gathered}$$

Step 5: Finding the maximum horizontal distance

As we know, the maximum horizontal distance (at $\theta =45°$) traveled by the body is,

$R_{max}=\frac{u^2}{g}$, which is equal to equation (1).

So, $R_{max}=3.47km$

Equation (1) says that the maximum horizontal distance traveled by the bullet is $3.47km$.

Hence, it is not possible for the bullet to hit a target 5 km away.