Course:MATH102/Question Challenge/2000 December Q8

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Question

A cannon-ball fired at ground level at angle $\theta$ to the horizon $(0\leq \theta \leq \pi /2)$ will travel a horizontal distance (called the range),$R$, given by the formula below:

$R={\frac {2}{g}}v_{0}^{2}\sin(\theta )\cos(\theta ).$

Here $v_{0}>0,$ the initial speed of the cannon-ball is a fixed constant, and $g=9.8{\mbox{m/s}}^{2}$ is acceleration due to gravity. Air resistance is neglected. (See diagram.) What is the maximum possible range?

The range $R$ for the cannon ball problem.

${\displaystyle }$

Hints

[show]Hint 1
It may simplify your work to use the trigonometric identity $2\sin(\theta )\cos(\theta )=\sin(2\theta )$

Solutions

[show]Solution

$R={\frac {2}{g}}v_{0}^{2}\sin \theta \cos \theta$ We will differentiate and set the equation equal to zero to solve for $\theta$: $R'={\frac {2}{g}}v_{0}^{2}(\sin \theta (-\sin \theta )+\cos \theta \cos \theta )$ $0=\cos ^{2}\theta -\sin ^{2}\theta$ $\cos ^{2}\theta =\sin ^{2}\theta$ ${\frac {\sin ^{2}\theta }{\cos ^{2}\theta }}=1$ $\tan ^{2}\theta =1$ $\theta ={\frac {\pi }{4}}$ = 45 degrees To find the maximum range we will substitute this value of $\theta$ into the function for the range of the cannonball: $R={\frac {2}{g}}v_{0}^{2}\sin {\frac {\pi }{4}}\cos {\frac {\pi }{4}}$ We know that this is a maximum range and not a minimum range because the other two critical points are global minimums. When the values in between these minimums are increasing, we know that the critical point in between them will produce a maximum.