Chapter 12.3, Problem 43E

To determine

To Prove: The shot will remain in the air for a total of $t=\frac{v_0\sin \theta +\sqrt{v_0{}^2{\sin }^2\theta +2gh}}{g}\text{ seconds}$ and will travel a horizontal distance of $\frac{v_0{}^2\cos \theta }{g}\left(sin\theta +\sqrt{{\sin }^2\theta +\frac{2gh}{v_0{}^2}}\right)\text{ feet}$ if the path of a shot thrown at an angle $\theta$ is $r\left(t\right)=\left(v_{0}vos\theta \right)ti+\left[h+\left(v_0\sin \theta \right)t-\frac{1}{2}g{t}^2\right]j$.