I'm teaching myself mechanics, and set out to solve a problem determining the optimum angle to throw a projectile when standing on a hill, for maximum range. My answer seems almost plausible, except for one term, which, to be plausible, needs to have its sign switched. But I can find no hole in my reasoning.
Problem: I am standing on a straight, downward sloped hill, and wish to throw a rock for maximum range. The hill is sloped down from horizontal by $\varphi$. What angle $\theta$ above the horizontal should I throw it at?
My solution:
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Use coordinates so that $x$ is parallel to the hill
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Let $\alpha = \varphi + \theta$ (that is, the angle above the ground that I'm throwing at)
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Then initial $v$ is $v_x = \cos \alpha$, $v_y = \sin \alpha$ (normalizing the units to remove any constants)
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Acceleration due to gravity is then $a_x = -k \sin \varphi$, $a_y = – k \cos \varphi$ (gravity is in the y direction). To make the calculations simpler, assume $k = 2$ (answer holds for any value of gravity, so is same on Moon as on Earth)
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We want to find the alpha which maximizes $s_x$ at the time that makes $s_y = 0$. First, find the time which makes $s_y = 0$; call it t.
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$s_y = t \sin \alpha – t^2 \cos \varphi$. Using the quadratic formula, $s_y = 0$ at $t = 0$ or $t = \frac{ \sin \alpha }{\cos \varphi}$.
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Now, find $s_x$ at this $t$. Substituting in and using basic algebra and trig, we get $s_x = \sin \alpha \, \cos \alpha – \sin ^2 \alpha\, \tan \varphi$. (This makes sense; the first term maxes at $\pi/4$, like we'd expect from symmetry. The second term tells us that if the ground banks down significantly, we should lower our angle of throwing. Very plausible.)
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Taking phi as a constant, we wish to maximize this expression. A little calculus and trig identities gets the derivative equal to $\cos(2\alpha)- \sin(2\alpha) \,\tan \varphi$, which has a zero at $\alpha = \pi/4 – \varphi/2$, or $\theta = \pi/4 – 3\varphi/2$. Here's where things break down. The first term, $\pi/4$, seems correct. But the second term gives ludicrous results.
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Switching the sign of the second term in the alpha equation ends up with $\theta = \pi/4 – \varphi/2$, which gives completely plausible results. But I can't find any error in my reasoning or calculations!
Can anyone find the missing link?
Answer explanation:
As Pygmalion determined, step 4 is wrong. The $a_y$ value is correct, but, $a_x$ should be positive: pointing down the hill.
The answer is independent of the magnitude of gravity; but it depends on the direction.
Revising the derivation:
7. $s_x = \sin \alpha \cos \alpha + \sin ^2 \alpha \tan \phi$
8. Derivative is $\cos(2 \alpha) + \sin(2 \alpha) \tan \phi$, with zero at $\alpha = \pi/4 + \phi/2$, thus $\theta = \pi/4 – \phi/2$. QED.
Best Answer
As far as I see, $a_y = - k \cos(\varphi) <0$, but $a_x = k \sin(\varphi) >0$! At least if you are throwing the projectile in the downward direction...