Although there’s an element of Calculus in this week’s post, and even more so, of Trigonometry, I am including it in the Algebra category because that is the more interesting aspect of this problem.
The following equation is the path of a projectile propelled at an angle \(\boldsymbol\theta\), where:
- \(\boldsymbol y\) is the height
- \(\boldsymbol x\) is the horizontal distance
- \(\boldsymbol g = 32\, ft/sec^2\) (acceleration due to gravity)
- \(\boldsymbol v_\boldsymbol 0 = 24\, ft/sec\) (initial velocity)
- \(\boldsymbol h = 9\, ft\) (initial height)
\[y = \frac{-g \left(\sec^{2}\theta \right) x^{2}}{2 \, v_{0}^{2}}+\left(\tan\theta\right) x + h, \qquad 0 \le \theta \le \frac{\pi}{2}\]
Neglecting air resistance, what value of \(\boldsymbol \theta\) will produce a maximum horizontal distance?
First, we substitute some of the variables with the values given above and set \(\boldsymbol y = 0\); this last is because we are interested in horizontal distance and not height, they are separate vectors. Next, we simplify.
\begin{alignat*}{1}
0 &= \frac{-32 \left(\sec^{2}\theta\right) x^{2}}{2 \left(24^2\right)}+\left(\tan\theta\right) x + 9\\
0 &= \frac{-\sec^{2}\theta}{36} x^{2} + \left(\tan\theta\right) x + 9
\end{alignat*}
This is the part of the problem I liked best. We solve for \(\boldsymbol x \) using ….. the Quadratic Formula, with \(\boldsymbol a = \frac{-\sec^{2}\theta}{36}, \; \boldsymbol b = \tan\theta\), \(\,\) and \(\boldsymbol c = 9\).
\begin{alignat*}{1}
x &= \frac{-b \pm \sqrt{b^{2} – 4 \, a \, c}}{2 \, a}\\
x &= \frac{-\tan\theta \pm \sqrt{\tan^{2}\theta -4 (\frac{-\sec^{2}\theta}{36}) (9)}}{2 ( \frac{-\sec^{2}\theta}{36})}
\end{alignat*}
It looks very messy, but it is surprisingly easy to simplify because most of the numeric values cancel each other and we can use the basic trig identities: \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), and \(\sec\theta = \frac{1}{\cos\theta}\).
\begin{alignat*}{1}
x &= \frac{- \tan\theta \pm \sqrt{\tan^{2}\theta + \sec^{2}\theta}}{\frac{-\sec^{2}\theta}{18}}\\
x &= 18 \left(\frac{\sin\theta}{\cos\theta}\right)\left(\cos^{2}\theta\right) + 18\left(\cos^{2}\theta\right) \sqrt{\frac{\sin^{2}\theta}{\cos^{2}\theta} + \frac{1}{\cos^{2}\theta}}
\end{alignat*}
Now, we cancel one of the cosines in the first term and, having a cosine squared as common denominator under the radical, we move cosine out from under and cancel it with one of the cosines in front of the radical.
\[x = 18 \left(\sin\theta\right) \left(\cos\theta\right) + 18 \left(\cos\theta\right) \sqrt{\sin^{2}\theta + 1}\]
Factoring is the last manipulation we do to this equation:
\[x = 18\left(\cos\theta\right) \left(\sin\theta + \sqrt{\sin^{2}\theta + 1}\right),\quad x \ge 0\]
We need to find the value of \(\boldsymbol \theta\) that gives us the maximum value of \(\boldsymbol x\) (horizontal distance). For this we differentiate and apply the First Derivative Test. Too messy to do it manually. Even the textbook advices to use technology for this task.
The result is that the maximum value of \(\boldsymbol x\) occurs when
\[\theta \approx 0.61548\,radian, \quad or \quad \theta \approx 35.3\,degrees\]
The longest of the three paths shown in this graph, is when the angle \(\boldsymbol \theta = 35\) degrees:
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected] .