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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:

  1. Use coordinates so that $x$ is parallel to the hill

  2. Let $\alpha = \varphi + \theta$ (that is, the angle above the ground that I'm throwing at)

  3. Then initial $v$ is $v_x = \cos \alpha$, $v_y = \sin \alpha$ (normalizing the units to remove any constants)

  4. 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)

  5. 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.

  6. $s_y = t \sin \alpha - t^2 \cos \varphi$. Using the quadratic formula, $s_y = 0$ at $t = 0$ or $t = \frac{ \sin \alpha }{\cos \varphi}$.

  7. 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.)

  8. 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.

  9. 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.

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Welcome to PhysSE. Please next time use the math formatting so x = sin theta becomes $x = \sin \theta$ for example. –  ja72 Apr 22 '12 at 21:36
    
One point: I claim the answer is independent of gravity, same on Moon and Mars. But what about "backwards gravity" (pushing away from ground)? That would surely change the answer. This doesn't solve things, but it may be a good start. –  S. Robert James Apr 22 '12 at 21:38
    
Why not keep the horizontal vertical coordinate system (for $x$, $y$ components) and match the trajectory $y(t)$,$x(t)$ to the downwards slope of $-\tan\varphi$ or $y(t)=-x(t)\,\tan\varphi$ ? –  ja72 Apr 23 '12 at 0:42
    
@S.RobertJames: I think that the answer is dependent upon the ratio of the initial speed to gravity. –  Manishearth Apr 23 '12 at 4:04
    
Alright, it's independent of $g$. My formula is $R=\frac{2u^2}{g\cos\varphi}\left(\cos\theta\sin(\varphi+\theta)\right)$, maximum at $\theta=\frac\pi2-\frac\varphi2$. What's yours? –  Manishearth Apr 23 '12 at 4:06
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1 Answer

up vote 1 down vote accepted

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...

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I believe they're both $< 0$. $a$ is due to gravity, and gravity is straight down, rotated clockwise by $\phi$ due to the coordinate rotation. Rotating $[0 -1]$ by $\phi$, $0 < \phi < \pi / 2$ yields both components negative. –  S. Robert James Apr 23 '12 at 5:49
    
@S.RobertJames If you are throwing downward, than maximum distance is greater, this is because $a_y$ is smaller and $a_x$ is positive (helping you). Forget rotating, draw the picture. –  Pygmalion Apr 23 '12 at 5:51
    
Please explain why you feel $a_x$ is positive (helping). Although I see your point intuitively (gravity helps me throw down a hill), doesn't rotating the coordinates (as described in my prev. comment) yield a negative (hurting) $a_x$? –  S. Robert James Apr 23 '12 at 5:54
    
I read the $0!$ as "zero-factorial". Made me go "wait, what?". Then I looked back at the question to make sure that its not a recursive one (where there are multiple bounces). ;-) –  Manishearth Apr 23 '12 at 5:55
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S. Robert James: Pleased to hell you. :) –  Pygmalion Apr 23 '12 at 6:01
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