Question

A projectile’s horizontal range over level ground is $\raisebox{1ex}{${v_0}^2\sin 2\theta $}\!\left/ \!\raisebox{-1ex}{$g$}\right.$. At what launch angle or angles will the projectile land at half of its maximum possible range?

Short answer

The required angles are$15°\text{and}75°$.

Step-by-step solution

Step 1. Given information

The formula to calculate the range$\left(R\right)$of a projectile is given by

localid="1648260841963" $R=\raisebox{1ex}{${v_0}^2\sin 2\theta $}\!\left/ \!\raisebox{-1ex}{$g$}\right..........................\left(1\right)$

Step 2. Calculation for the condition of maximum range

The range of the projectile will be maximum when

$\sin 2\theta =1......................\left(2\right)$

Solve equation (2) to calculate the required projection angle for which the range is maximum.

$\begin{array}{rcl}\sin 2\theta & =& 1\\ & =& \sin 90°\\ 2\theta & =& 90°\\ \theta & =& 45°\end{array}$

Since $\sin \left(180°-\theta \right)=\sin \theta$, the projectile will cover the maximum range for the projection angle being$\theta$or$\left(90°-\theta \right)$.

Step 3. Calculation for the condition of half the maximum range 

The projectile will cover half the maximum range when

$\sin 2\theta =\frac{1}{2}................\left(3\right)$

Solve equation (3) to calculate one of the required projection angle.

$\begin{array}{rcl}\sin 2\theta & =& \frac{1}{2}\\ & =& \sin 30°\\ 2\theta & =& 30°\\ \theta & =& 15°\end{array}$

The other value of the required projection angle is given by

$\begin{array}{rcl}\left(90°-\theta \right)& =& 90°-15°\\ & =& 75°\end{array}$