Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The question what-is-the-optimum-angle-of-projection-when-throwing-a-stone-off-a-cliff was asked and answered a while back. This one has a much cleaner answer. Now you are on a uniform slope, a line through the origin that is not horizontal and you want to throw a stone as far as possible. As a function of the angle of the slope, what angle should you throw at? Usual assumptions: uniform gravity, no friction.

share|improve this question
    
Are you asking this question or just setting it as a challenge? –  anon Oct 21 '10 at 23:11
1  
Just setting a challenge. I remember the answer. –  Ross Millikan Oct 21 '10 at 23:15

2 Answers 2

up vote 6 down vote accepted

I am surprised no one seems to have answered this.

I get $$\pi/4 + \alpha/2$$.

If the line is given by $y = x \tan \alpha$, with $\alpha$ acute and we throw from the origin at an angle $\theta$ from the x-axis, at velocity $1$, then we have that the projectile satisfies, assuming gravity $g=2$ (in appropriate units)

$\displaystyle y = t\sin\theta - t^2$, $\displaystyle x = t\cos \theta$

The time at which it intersects the line again is given by

$\displaystyle t\sin\theta - t^2 = t \tan\alpha \cos \theta$ and so

$\displaystyle t = \sin \theta - \tan \alpha \cos \theta$

It is enough to maximize the horizontal distance travelled by the projectile, which is given by

$\displaystyle \cos \theta (\sin \theta - \tan \alpha \cos \theta)$

With little manipulation, we need to maximize

$\displaystyle \sin(2\theta - \alpha)$

which gives the result.

share|improve this answer
    
That's what I got. –  Ross Millikan Oct 22 '10 at 19:28

When the solution is so neat, I feel like it should have a high-level explanation "from the book" that makes it immediately obvious. Sadly, I was not able to find one, but I'm posting what I did get in hopes that it might inspire someone else.

Imagine firing projectiles at all angles from the origin. Assume unit speed and gravity $g = 2$, as in Moron's answer. Consider a frame of reference which starts at the origin and falls freely. There is no gravity in this frame of reference, so all projectiles expand radially outward forming a cone, $$x^2 + y^2 = t^{2}.$$ However, the ground is no longer given by $y = x \tan\alpha$ but rather by $$y = x\tan\alpha + t^{2},$$ which is a parabolic cylinder. The projectiles hit the ground at the points where the cone and the parabolic cylinder intersect. Subtracting the two equations, we find that this intersection satisfies $$\left(x^2 + x\tan\alpha\right) + \left(y^2 - y\right) = 0,$$ i.e. on the $xy$ plane it projects to a circle. On physical grounds, one can see that the circle passes through the origin, and is tangent to the line $y = x\tan\alpha$ there. The projectile that reaches the farthest corresponds to the point with the largest $x$-coordinate, and the angle of the projectile is simply the angle of the line joining that point to the origin (because projectiles just move radially outward in this frame of reference). I encourage you to draw a little diagram and see that this line bisects the angle between the vertical, at $\pi/2$, and the ground, at $\alpha$.

share|improve this answer
    
You sound like a junior Einstein thinking about this the falling frame. Thanks. –  Ross Millikan Oct 23 '10 at 15:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.