Chapter 12.3, Problem 41E

Shot-Put Throw The path of a shot thrown at an angle $\theta$ is

$r(t)=(v_0\cos \theta )ti+\left[h+(v_0\sin \theta )t-\frac{1}{2}g{t}^2\right]j$

where $v_0$ is the initial speed, h is the initial height, t is the time in seconds, and g is the acceleration due to gravity. Verify that 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

$\frac{v_0^2\cos \theta }{g}\left(\text{sin}\theta +\sqrt{{\sin }^2\theta +\frac{2gh}{v_0^2}}\right)\text{feet}\text{.}$