Projectile Motion

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 \[ \begin{eqnarray*} \dfrac{dy}{dx} =\dfrac{v_{0}\sin \theta -gt}{v_{0}\cos \theta } &=& 0 \notag\\ v_{0}\sin \theta -gt &=&0 \notag \\ t &=&\dfrac{v_{0}\sin \theta }{g} \end{eqnarray*} \]