You are target shooting using a toy gun that fires a small ball at a speed of 15 m/s. When the gun is fired at an angle of 30° above horizontal, the ball hits the bull’s-eye of a target at the same height as the gun. Then the target distance is halved. At what angle must you aim the gun to hit the bull’s-eye in its new position? (Mathematically there are two solutions to this problem; the physically reasonable answer is the smaller of the two.)

The initial speed of the ball is $15\text{m/s}$and the projection angle is $30°$,.

The formula to calculate the range of the ball is given by

$R=\frac{{v_0}^2\sin 2\theta }{g}.............\left(1\right)$

Here, $R,v_0,\theta ,g$are the range, initial speed of the ball, projection angle and acceleration due to gravity respectively.

Substitute $15\text{m/s}$for $v_0$, $30°$for $\theta$and $9.80{\text{m/s}}^2$for $g$into equation (1) to calculate the required range of the ball.

$\begin{array}{rcl}R& =& \frac{{\left(15\text{m/s}\right)}^2\times \sin \left(2\times 30°\right)}{9.80{\text{m/s}}^2}\\ & \approx & 20\text{m}\end{array}$

Let us assume that when the target distance is halved, the new projection angle is $\theta \text{'}$.

The formula to calculate the new projection angle is given by

$\frac{R}{2}=\frac{{v_0}^2\sin 2\theta \text{'}}{g}.....................\left(2\right)$

Substitute the given values of the known parameters into equation (2) and solve to calculate the required projection angle.

$\begin{array}{rcl}\frac{20\text{m}}{2}& =& \frac{{\left(15\text{m/s}\right)}^2\times \sin 2\theta \text{'}}{9.80{\text{m/s}}^2}\\ \sin 2\theta \text{'}& =& \frac{10\text{m}\times 9.80{\text{m/s}}^2}{{\left(15\text{m/s}\right)}^2}\\ & \approx & 0.435\\ 2\theta \text{'}& =& {\sin }^{-1}\left(0.435\right)\\ & \approx & 26°\\ \theta \text{'}& =& 13°\end{array}$