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Problem: Show that if a projectile is shot from a height $h$ with speed $v_0$ the maximum range obtains for launch angle $\theta = \arctan(\frac{v_{0}} {\sqrt{2gh+v_{0}^{2}}})$.

I'm able to derive the general equation for range (i.e. $y \neq0$) without any bother, but I get all tangled up in the differentiation and trig once I try to maximise $\theta$. If someone could guide me through the derivation, that would be fantastic:)

My progress thus far:

We know that for a projectile fired from initial height $h$ with initial velocity $v_{0}$ at an angle $\theta$, the equations of motion are

$x(t)= v_{0}cos{\theta}t$; $y(t)=h + v_{0}sin{\theta}t - \frac{1}{2}gt^{2}$

Range occurs at the time of flight ($t^{*})$, which is found by setting the equation for $y(t)$ equal to zero. Doing so yields

$h + v_{0}sin{\theta}t^{*}-\frac{1}{2}gt^{*2}$ = 0

Rearranging and solving for $t^{*}$:

$t^{*}=\frac{-(-2v_{0}sin\theta)\pm\sqrt{(-2v_{0}^{2}sin^{2}\theta-2(g)(-2h)}}{2g}$

$=\frac{v_{0}sin\theta}{g}\pm \frac{1}{g}\sqrt{v_{0}^{2}sin^{2}\theta+2gh}$

The square-root must be positive, and since the initial velocity and sine of the launch angle can also be assumed to be positive, it follows that the greater time will occur when using the plus sign from the plus or minus sign.

$t^{*}=\frac{v_{0}sin\theta}{g}+ \frac{1}{g}\sqrt{v_{0}^{2}sin^{2}\theta+2gh}$

Substituting into our equation for $x(t)$, we find that the general expression for the range $R$ is given by

$x(t^{*})=R=v_{0}cos\theta[\frac{v_{0}sin\theta}{g}+ \frac{1}{g}\sqrt{v_{0}^{2}sin^{2}\theta+2gh}]$

or

$R = \frac{v_{0}cos\theta}{g}[v_{0}sin\theta+\sqrt{v_{0}^{2}sin^{2}\theta+2gh}]$

From here, I know to set $dR/d\theta$ equal to 0, but I keep fumbling the differentiation.

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  • $\begingroup$ Can you show what you have done so far? $\endgroup$ – relayman357 Sep 7 '20 at 12:13
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There is an alternative approach to this problem which avoids the need for calculus. Solve your x equation for cos(θ) and the y equation for sin(θ). Then $sin^2$(θ) + $cos^2$(θ) = 1. This leads to a quadratic equation in $t^2$ which gives two positive values for t. These correspond to the two possible angles for hitting a target a known distance down range. As you approach the maximum range, the angles (and times) converge. At this point the square root in the quadratic equals zero. You can set it equal to zero and solve for $x^2$. With the square root at zero the quadratic gives $t^2$. Combine these in the x equation to get cos(θ). So far, I have not been able to combine these to get the formula you were given for tan(θ). However, I have a spreadsheet which solves numeric projectile problems. For this one I had to use a tool called solver. For several different chosen sets of numbers, my formulas (and yours) agree with the predictions on the spreadsheet. (If I start feeling ambitious, I may tackle the derivative which has you blocked.)

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  • $\begingroup$ R.W. Bird, Solver is a wonderful tool (as I'm sure that you know). Also, that derivative looks like it's going to be challenging if you don't use software to work it. Good luck. $\endgroup$ – David White Sep 9 '20 at 18:22

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