A projectile is fired at an angle $\theta ,$ $0<\theta <\dfrac{\pi }{2},$ to the horizontal with an initial speed of $v_{0}$ m/s. Assuming no air resistance, the position of the projectile after $t$ seconds is given by the parametric equations $x(t) =( v_{0}\cos \theta ) t,$ $y(t) =( v_{0}\sin \theta ) t-\dfrac{1}{2} gt^{2},\quad t\geq 0,$ where $g$ is the acceleration due to gravity.

(a) Find the slope of the tangent line to the motion of the projectile as a function of $t$.

(b) At what time is the projectile at its maximum height?

Solution (a) The slope of the tangent line is given by $\dfrac{dy }{dx}.$ $$\begin{equation*} \dfrac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{d}{ dt}\left[ \left( v_{0}\sin \theta \right) t-\dfrac{1}{2}gt^{2}\right] }{ \dfrac{d}{dt}\left[ \left( v_{0}\cos \theta \right) t\right] }=\dfrac{ v_{0}\sin \theta -gt}{v_{0}\cos \theta }=\tan \theta -\dfrac{gt}{v_{0}\cos \theta } \end{equation*}$$

(b) The projectile is at its maximum height when the slope of the tangent line equals $0.$ That is, when $$\begin{eqnarray*} \dfrac{dy}{dx} =\dfrac{v_{0}\sin \theta -gt}{v_{0}\cos \theta } &=& 0 \notag\\ v_{0}\sin \theta -gt &=&0 \notag \\ t &=&\dfrac{v_{0}\sin \theta }{g} \end{eqnarray*}$$