# What is the optimal angle when throwing a stone on a slope?

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.

• Are you asking this question or just setting it as a challenge? – anon Oct 21 '10 at 23:11
• Just setting a challenge. I remember the answer. – Ross Millikan Oct 21 '10 at 23:15
• Should we take into account anatomic concerns? Because for example to throw a stone at a 45° angle, one have to release it quite early. Which would cause a lower speed compared to a later release on a lower angle. So we might have a speed/angle trade-off here. Also, the arm and wrist configuration should cause a non linear acceleration curve. (peaking at lower angles?) – tuxayo Aug 12 '17 at 2:17

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.

• 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 $\alpha$.

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