Figure 43

An object is propelled from ground level at an angle $\theta$, $0\lt \theta \lt \dfrac{\pi}{2}$, to the horizontal with an initial velocity of 16feet/second, as shown in Figure 43. The equations of the horizontal position $x = x(\theta)$ and the vertical position $y = y(\theta)$ of the object after t seconds are given by $$\begin{equation*} x = x(\theta) = (16\cos \theta) t\qquad \hbox{and}\qquad y = y(\theta) = -16t^{2}+(16\sin \theta) t \end{equation*}$$

For a fixed time t:

Solution (a) $$\begin{eqnarray*} \lim\limits_{\theta \rightarrow 0^{+}}x(\theta) &=&\Big[\lim\limits_{\theta \rightarrow 0^{+}}(16\cos \theta)\Big] t = 16t\qquad{\color{#0066A7}{\hbox{$t$ fixed}}}\\[3pt] \lim\limits_{\theta \rightarrow 0^{+}}y(\theta) &=&\lim\limits_{\theta \rightarrow 0^{+}} [ -16t^{2}+(16\sin \theta) t ]\\[3pt] &=&\lim\limits_{\theta \rightarrow 0^{+}}(-16t^{2}) +\Big[\lim\limits_{\theta \rightarrow 0^{+}}(16\sin \theta)\Big] t = -16t^{2} \end{eqnarray*}$$

(b) The limits in (a) tell us that the horizontal position x of the object moves to the right ($x = 16t$) over time and that the vertical position y of the object immediately moves downward ($y = -16t^{2}$) into the ground. Neither of these conclusions is consistent with what we expect from the motion of an object propelled at an angle close to $\theta = 0$.

(c) $$\begin{eqnarray*} \lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}x(\theta) &=&\Big[\lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}(16\cos \theta)\Big] t = 0\\[3pt] \lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}y(\theta) &=&\lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}\left[ -16t^{2}+(16\sin \theta) t\right] = \lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}(-16t^{2}) +\Big[\lim\limits_{\theta \rightarrow \frac{\pi}{2}^{-}}(16\sin \theta)\Big] t\\[3pt] &=&-16t^{2}+16t \end{eqnarray*}$$

(d) The limits we found in (c) tell us that the horizontal position remains at $0$. The vertical position $y = -16t^{2}+16t = -16t(t-1)$ tells us the object has a maximum height of 4feet and returns to the ground after 1 second. (The parabola $y = -16t^{2}+16t$ opens down and has a maximum value at $t = \dfrac{-b}{2a} = \dfrac{-16}{-32} = \dfrac{1}{2}$ at which time $y = 4$.) These conclusions are consistent with what we would expect from the motion of an object propelled close to the vertical.