Optimizing Projectile Range on Uniformly Sloped Surfaces

[thrill3rnit3]

Homework Statement

Hello. I'm doing problems out of Kleppner and Kolenkow and for some reason I'm stuck at one of them.

A boy stands A boy stands at the peak of a hill which slopes downward uniformly at angle \phi. At what angle \theta from the horizontal should he throw a rock so that it has the greatest range?

here's an image:

http://img42.imageshack.us/img42/4135/phys.gif

Homework Equations

equations of motion

The Attempt at a Solution

Here's the work I had on a sheet of paper:

http://img843.imageshack.us/img843/2717/physa.jpg

The last equation on it should have a 2 in front of the Vo^2/g, but after that I'm stuck.

 

[Delphi51]

I don't have the solution; hope others will help!
Your work looks good (with the 2 added).
The next step would be to differentiate and set dx/dt = 0 to find the condition for maximum x. Unfortunately I get a trigonometric equation that I can't solve.

There is quite a writeup of this problem on Wikipedia, where the contributor finds the maximum "range along the slope" which I expect is the same as the maximum horizontal range. Very complicated, but it simplifies in the end and it should be easy to find the angle. See http://en.wikipedia.org/wiki/Trajectory

 

[mugaliens]

I'll give you a hint: It's not the same angle as works for best horizontal range. :) That would have been 45 deg assuming atmospheric effects are negligible. Second hint: Due to the downward slope, the horizontal component becomes somewhat more important than the vertical component. Third hint: Consider using trig equalities t reduce your terms. Fourth hint...

Let's so how you do given the above, first.

 

[thrill3rnit3]

So it all boils down to math and trig identities?

 

[Delphi51]

There is quite a writeup of this problem on Wikipedia, where the contributor finds the maximum "range along the slope" which I expect is the same as the maximum horizontal range.

I see I did not say this correctly. I meant that the angle that maximizes the range along the slope will also maximize the "x range" you have in your original post. This must be true since the range along the slope is x/cos(Φ).

But the direct approach of maximizing x that you started now looks easier. The trig equation from dx/dθ = 0 is much simplified by using a couple of double angle identities for sin(2θ) and cos(2θ).

 

[thrill3rnit3]

Why would it be dx/dθ instead of dx/dt as you said in your 1st post?

 

[Delphi51]

I seem to be making mistakes! Should be dx/dθ in both places. We are trying to find the value of θ that makes x a maximum so we are interested in how x varies with θ and that's why you used
t = x/(v*cosθ) to eliminate t. dx/dθ = 0 at the maximum.

 

[thrill3rnit3]

I seem to be making mistakes! Should be dx/dθ in both places. We are trying to find the value of θ that makes x a maximum so we are interested in how x varies with θ and that's why you used
t = x/(v*cosθ) to eliminate t. dx/dθ = 0 at the maximum.

yeah that's what I originally thought then I saw your dx/dt I was even more confused

It's all good though. Let me work on it a for a while...

 

[thrill3rnit3]

Thank you very much

 

[Delphi51]

Most welcome! Check to make sure that when Φ = 0 (or in the limit as Φ approaches 0) it says the optimal angle is 45 degrees. I seem to have an extra minus sign in my solution.

 

[thrill3rnit3]

Most welcome! Check to make sure that when Φ = 0 (or in the limit as Φ approaches 0) it says the optimal angle is 45 degrees. I seem to have an extra minus sign in my solution.

Mine too has the negative sign. I guess because Φ is an angle of depression (negative angle?) everything works out?

 

[Delphi51]

Ah, perhaps. Good to hear it worked out for you!