Question

A projectile is launched from ground level at angle $\theta$and speed

$v_0$into a headwind that causes a constant horizontal acceleration

of magnitude $a$opposite the direction of motion.

a. Find an expression in terms of $a$and $g$for the launch angle

that gives maximum range.

b. What is the angle for maximum range if $a$is 10% of $g$?

Short answer

(a) the launch angle is given by $\theta =\frac{\pi }{4}-\frac{1}{2}{\tan }^{-1}\left(\frac{a}{g}\right)$

(b) the required value of the angle is$\begin{array}{rcl}& & 39.3°\end{array}$.

Step-by-step solution

Part (a): Step 1. Given information

The initial speed of the projectile is$v_0$and the projection angle is$\theta$.

Part (a): Step 2. Calculation for the time taken by the projectile to cover the maximum range

The formula to calculate the vertical height travelled by the projectile is given by

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

Here, $y$is the vertical distance and $t$is the time taken by the object to cover the maximum range.

Substitute $0$for $y$into equation (1) and solve to calculate the required time taken by the projectile.

$\begin{array}{rcl}0& =& v_0\sin \theta t-\frac{1}{2}g{t}^2\\ t& =& \frac{2v_0\sin \theta }{g}........................\left(2\right)\end{array}$

Part (a): Step 3. Calculation of the maximum range

The formula to calculate the horizontal distance $\left(R\right)$travelled by the projectile is given by

role="math" localid="1647405477077" $R=v_0\cos \theta t-\frac{1}{2}a{t}^2.................\left(3\right)$

Substitute the value of time from equation (2) into equation (3) and solve to obtain the horizontal distance travelled.

role="math" localid="1647405893350" $\begin{array}{rcl}R& =& v_0\cos \theta \left(\frac{2v_0\sin \theta }{g}\right)-\frac{1}{2}a{\left(\frac{2v_0\sin \theta }{g}\right)}^2\\ & =& \frac{{v_0}^2}{g}\left(\sin 2\theta -\frac{a}{g}2{\sin }^2\theta \right)\\ & =& \frac{{v_0}^2}{g}\left(\sin 2\theta -\frac{a}{g}\left(1-\cos 2\theta \right)\right)\\ & =& \frac{{v_0}^2}{g}\left(\sin 2\theta +\frac{a}{g}\cos 2\theta \right)-\frac{{v_0}^2}{g^2}a\\ & =& \frac{{v_0}^2}{g^2}\sqrt{a^2+g^2}\left(\frac{g}{\sqrt{a^2+g^2}}\sin 2\theta +\frac{a}{\sqrt{a^2+g^2}}\cos 2\theta \right)-\frac{{v_0}^2}{g^2}a.............\left(4\right)\end{array}$

Substitute $\tan \alpha =\frac{a}{g}$into equation (4) and simplify to obtain the range.

$\begin{array}{rcl}R& =& \frac{{v_0}^2}{g^2}\sqrt{a^2+g^2}\left(\frac{g}{\sqrt{a^2+g^2}}\sin 2\theta +\frac{a}{\sqrt{a^2+g^2}}\cos 2\theta \right)-\frac{{v_0}^2}{g^2}a\\ & =& \frac{{v_0}^2}{g^2}\sqrt{a^2+g^2}\sin \left(2\theta +\alpha \right)-\frac{{v_0}^2}{g^2}a......................\left(5\right)\end{array}$

Part (a): Step 4. Calculation of the launch angle

The range of the projectile is maximum when

$\sin \left(2\theta +\alpha \right)=1...................\left(6\right)$

Solve equation (6) to obtain the required launch angle.

$\begin{array}{rcl}\sin \left(2\theta +\alpha \right)& =& 1\\ & =& \sin \frac{\pi }{2}\\ 2\theta +\alpha & =& \frac{\pi }{2}\\ \theta & =& \frac{1}{2}\left(\frac{\pi }{2}-\alpha \right)\\ & =& \frac{\pi }{4}-{\tan }^{-1}\left(\frac{a}{g}\right).........................\left(7\right)\end{array}$

Part (b): Given information

The acceleration$\left(a\right)$ opposing the horizontal motion of the projectile is$10\%$of$g$.

Part (b): Step 2. Calculation of the launch angle

Substitute $10\%g$for $a$into equation (7) and solve to calculate the required launch angle.

$\begin{array}{rcl}\theta & =& \frac{\pi }{4}-{\tan }^{-1}\left(\frac{10\%g}{g}\right)\\ & \approx & 39.3°\end{array}$