## Trigonometry (11th Edition) Clone

An angle of $41^{\circ}$ will maximize the distance.
We can find the angle $\theta$ which will maximize the distance: $\theta = arcsin \left(\sqrt{\frac{v^2}{2v^2+64h}}\right)$ $\theta = arcsin \left(\sqrt{\frac{(32)^2}{(2)(32)^2+(64)(5.0)}}\right)$ $\theta = arcsin \left(\sqrt{\frac{1024}{2048+320}}\right)$ $\theta = arcsin \left(\sqrt{\frac{1024}{2368}}\right)$ $\theta = arcsin \left(\sqrt{0.432432}\right)$ $\theta = arcsin \left(0.657596\right)$ $\theta = 41^{\circ}$ An angle of $41^{\circ}$ will maximize the distance.