Optimal launch angle for maximum range

Differentiate $R(\theta) = \frac{v^2}{g}\sin(2\theta)$ with respect to $\theta$, using the chain rule: $R'(\theta) = \frac{v^2}{g}\cdot 2\cos(2\theta) = \frac{2v^2}{g}\cos(2\theta)$. Setting $R'(\theta) = 0$ gives $\cos(2\theta) = 0$, so $2\theta = \pi/2$, which means $\theta = \pi/4$ is the only critical point in $(0, \pi/2)$.

Since $R(0) = 0$ and $R(\pi/2) = 0$ at the endpoints, while $R(\pi/4) = \frac{v^2}{g}\sin(\pi/2) = \frac{v^2}{g} > 0$, the closed interval method confirms $\theta = \pi/4$ gives the maximum range. This matches physical intuition: a 45° launch angle maximizes horizontal distance for a projectile with no air resistance.

The range equals zero at the two endpoints of the interval, namely angle zero and angle ninety degrees (pi over two radians). At angle zero the ball is launched flat along the ground, and at ninety degrees it is launched straight up, so in both cases the horizontal distance traveled is zero. This matches the closed interval method comparison used in part a, where the endpoint values were shown to be zero while the midpoint angle of forty-five degrees gave the maximum.