Maximize Range of Dropped Ball from a Wedge

[IHateMayonnaise]

Homework Statement

I'm attaching a drawing depicting the problem, hopefully the mods will approve it soon. In words: A ball is dropped from some height $h_0$ and hits a point on a wedge at some angle $\theta$ at some height up the wedge $h^\prime$. Optimize $h^\prime$ and $\theta$ for which the range is maximized.

Homework Equations

The Attempt at a Solution

My first reaction is to use calculus of variations, however I'm not quite sure this is a viable option, as I don't really see any constraints that I could use to create some Lagrange multipliers. The problem basically boils down to "if someone throws a ball at an angle $\theta$ with an initial velocity $v_0$ while standing on a cliff of height $h^\prime$, find the max range."

The velocity can be found by conservations of energy:

$$mg(h_0-h^\prime) = 1/2 m v^2 \rightarrow v = \sqrt{2g(h_0-h^\prime)}$$

Applying kinematics along the y-direction to get the time, then doing kinematics along the x direction I arrive at (haven't check this, but this isn't the problem):

$$\Delta x = \frac{\sqrt{2g(h_0-h^\prime)}\cos\theta}{g} \left[\sqrt{2g(h_0-h^\prime)}\sin\theta - \sqrt{2gh_0\sin^2\theta + 2gh^\prime\cos^2\theta}\right]$$

Assuming this is correct, and I'm not saying it is, what would I do? How would I go about optimizing this? I am looking to find/optimize both $h^\prime$ and $\theta$. Would I take the derivative with respect to $h^\prime$ and set it equal to zero to find the best value? Would there be an easier way?

Thanks yall

IHateMayonnaise

 

[hotvette]

Assuming your expression for $\Delta x$ is correct, the way to proceed is to take partial derivatives with respect to h' and $\theta$ and set each equal to zero. You will thus have two (nonlinear) equations in two unknowns to solve.

[edit] I think your expression for $\Delta x$ isn't correct. If you let h'/h₀ = a and factor h₀ from the right side (in brackets), the value of the right side in brackets is negative for all values of a between 0 and 1 for a reasonable value for $\theta$ (e.g. 45 degrees).

 

[IHateMayonnaise]

Assuming your expression for $\Delta x$ is correct, the way to proceed is to take partial derivatives with respect to h' and $\theta$ and set each equal to zero. You will thus have two (nonlinear) equations in two unknowns to solve.

Thanks for the reply hotvette.

I was afraid you were going to say that; as you can see this expression is not so fun, differentiably speaking. I checked my work and there were a couple mistakes, but it is equally disgusting. I differentiated in Mathematica, set them equal, and it can't solve it. My thoughts: I must be doing this wrong. I am fairly sure that there must be an analytical solution to this, and I would hate to have to have a numerical one...

 

[viscousflow]

Have you tried using conservation of momentum to find the angle $$\theta$$? There are also some things you can assume about it to give maximum range.

 

[IHateMayonnaise]

Have you tried using conservation of momentum to find the angle $$\theta$$? There are also some things you can assume about it to give maximum range.

I don't mean to be a lazy moocher, but can you be a bit more specific?

 

[viscousflow]

Correct me if I'm wrong, but if you assume no drag and only gravitational effects, from Dynamics an object shot in the air at an angle has a maximum range when the angle is 45 degrees. If you include drag effects its somewhere between 20 and 30 degrees.

Also, you must be able to use conservation of momentum with 2Dof to find an expression for $$\theta$$ to optimize $$h'$$.

 

[hotvette]

I think viscousflow makes a good point about the optimum angle being 45 degrees, but that would be true only if the angle doesn't also affect h'. In other words, if the ramp were fixed horizontally and hinged, different angles would affect h'. But, if the ramp can also be moved right/left, then h' wouldn't depend on the angle. In the end I think you may have a single variable optimization problem.