Question

A bullet fired at an angle of $30$ degrees with the horizontal hits the ground $3km$ away. By adjusting its angle of projection, can one hope to hit a target $5km$ away? Assume the muzzle speed to be fixed, and neglect air resistance.

Answer 1

Step 1: Given Data

Range of the projectile, $R=3km$

The angle of projection, $\theta =30°$

Step 2: Formula Used

The horizontal range for the projectile is given by, $R=\frac{{u_{\circ }}^2\sin \left(2\theta \right)}{g}$ …$\left[1\right]$

where $R$ is the range of projectile

$u_{\circ }$ is the initial projectile velocity

$g$ is the acceleration due to gravity

$\theta$ is the angle of the projectile

Step 3: Calculate the initial velocity

Put the given values in the equation $\left[1\right]$ to find $u_{\circ }$

$3=\frac{{u_{\circ }}^2\sin \left(2\times 30\right)}{g}$

$\Rightarrow {u_{\circ }}^2=\frac{3g}{\sin \left(60\right)}$

$\Rightarrow {u_{\circ }}^2=\frac{3g}{\frac{\sqrt{3}}{2}}$

$\Rightarrow {u_{\circ }}^2=2\sqrt{3}g$

$\Rightarrow u_{\circ }={\left(2\sqrt{3}g\right)}^{\frac{1}{2}}$

Step 4: Calculate the maximum range

The maximum horizontal range is achieved when it is fired at an angle of $45°$ with the horizontal.

$R_{max}=\frac{{u_{\circ }}^2\sin \left(2\times 45\right)}{g}$

$=\frac{{\left({\left(2\sqrt{3}g\right)}^{{\displaystyle \frac{1}{2}}}\right)}^2\sin \left(90\right)}{g}$

$=2\sqrt{3}$

$=3.46km$

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