A cannon whose muzzle is tilted upward at an angle of 30° shoots a ball at an initial velocity of 600 m/sec.
(a) Find the position vector at all times t >= 0.
(b) How much time does the ball spend in the air?
(c) How far does the cannonball travel?
(d) How high does the ball get?
(e) How far from the cannon does the ball land?
(f) What is the speed of the ball at the time of impact with the earth? Assume that the mouth of the cannon is at ground level and ignore air resistance.
(a) From Example 2.1.6 we find, resolving the initial velocity vector into its vertical and horizontal components, that \mathbf{v}_{0}=300 \sqrt{3} \mathbf{i}+300 \mathbf{j}. Moreover, if we place the origin so that it coincides with the mouth of the cannon, then the initial position vector \mathrm{f}_{0} is (0,0) = 0\mathbf{i}+0 \mathbf{j}. From (9) \mathbf{f}(t)=-\frac{1}{2} g t^{2} \mathbf{j}+\mathbf{v}_{0} t+\mathbf{f}_{0}
\begin{aligned}\mathbf{f}(t) &=-\frac{1}{2} g t^{2} \mathbf{j}+300 \sqrt{3} t \mathbf{i}+300 t \mathbf{j}+\mathbf{0} \\&=300 \sqrt{3} t \mathbf{i}+\left(300 t-\frac{1}{2} g t^{2}\right) \mathbf{j}\end{aligned}
(b) The ball hits the ground when the vertical component of the position vector is zero, that is, when 300 t-\frac{1}{2} g t^{2}=0, which occurs at t=600 / g seconds.
(c) The total distance traveled is
s=\int_{0}^{600 / g}\left(\frac{d s}{d t}\right) d t.
But
\begin{aligned}\frac{d s}{d t} &=v(t)=|\mathbf{v}(t)|=\left|\mathbf{f}^{\prime}(t)\right|=|300 \sqrt{3} \mathbf{i}+(300-g t) \mathbf{j}| \\&=\sqrt{270,000+(300-g t)^{2}}\end{aligned}Thus using formula (2.5.4), we have
\begin{aligned}s\left(t_{1}\right) &=\text { length of arc from} t_{0} \text { to } t_{1} \\&=\int_{t_{0}}^{t_{1}}\left(\frac{d s}{d t}\right) d t=\int_{t_{0}}^{t_{1}} \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t .\end{aligned}s=\int_{0}^{600 / 8}\left(\frac{d s}{d t}\right) d t=\int_{0}^{600 / 8} \sqrt{270,000+(300-g t)^{2}} d t \text { meters }
Let 300-g t=\sqrt{270,000} \tan \theta=300 \sqrt{3} \tan \theta. Then
-g d t=300 \sqrt{3} \sec ^{2} \theta d \theta
and
Also,
\sqrt{270,000+(300-g t)^{2}}=\sqrt{270,000\left(1+\tan ^{2} \theta\right)}=300 \sqrt{3} \sec \theta \text {. }When t=0,300=300 \sqrt{3} \tan \theta, \tan \theta=1 / \sqrt{3}, and \theta=\pi / 6. When t=600 / g,
\begin{aligned}\tan \theta &=-1 / \sqrt{3} \text { and } \theta=-\pi / 6 . \text { Thus } \\s &=-\frac{1}{g} \int_{\pi / 6}^{-\pi / 6}(300 \sqrt{3})(300 \sqrt{3}) \sec ^{3} \theta d \theta \\&=\frac{270,000}{g} \int_{-\pi / 6}^{\pi / 6} \sec ^{3} \theta d \theta=\frac{540,000}{g} \int_{0}^{\pi / 6} \sec ^{3} \theta d \theta \\&=\left.\frac{270,000}{g}(\ln |\sec \theta+\tan \theta|+\sec \theta \tan \theta)\right|_{0} ^{\pi / 6} \\&=\frac{270,000}{g}\left(\ln \left|\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right|+\frac{2}{\sqrt{3}} \frac{1}{\sqrt{3}}\right)=\frac{270,000}{g}\left(\ln \sqrt{3}+\frac{2}{3}\right) \approx 33.467 \mathrm{~km} .\end{aligned}
(d) The maximum height is achieved when d y / d t=0. That is, when 300-g t=0 or when t=300 / g. For that value of t, y_{\max }=45,000 / g \approx 4587.2 \mathrm{~m}=4.5872 \mathrm{~km}.
(e) In 600 / g seconds, the x-component of f increases from 0 to (300 \sqrt{3}) \cdot(600 / g)\approx 31,780.7 \mathrm{~m} \approx 31.78 \mathrm{~km}.
(f) Since speed =v(t)=\sqrt{270,000+(300-g t)^{2}} meters per second, we find that upon impact t=600 / g seconds, so that
\begin{aligned}v\left(\frac{600}{g}\right) &=\sqrt{270,000+\left(300-g \cdot \frac{600}{g}\right)^{2}}=\sqrt{270,000+90,000} \\&=\sqrt{360,000}=600 \mathrm{~m} / \mathrm{sec}\end{aligned}These results are illustrated in Figure 2.
