Slanted Gravity: A boy on a hill

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SUMMARY

The discussion focuses on determining the optimal angle \(\theta\) from the horizontal for a boy to throw a rock from a hill sloping at angle \(\phi\) to achieve maximum range. The user employs a method involving the decomposition of vectors and kinematic equations, ultimately suggesting that \(\theta = \frac{90 - \phi}{2}\) maximizes the distance. The approach is validated by a forum member who confirms the method's validity while suggesting a more precise mathematical language. The conversation also seeks alternative, more elegant solutions to the problem.

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Homework Statement


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 is has the greatest range.

(Source: Introduction to Mechanics by Kleppner and Kolenkow)

Homework Equations


I suppose trig identities, vectors, and the general one dimensional kinematics would help (x_0+v_0t+a_0t^2)

The Attempt at a Solution


Now my solution went like this: Pretend the ground is flat and that instead gravity is at an angle \phi-90), and that the ball is launch at angle \theta+\phi My idea was to decompose both the gravity vector and the vector of the launch into vertical and horizontal vectors, look at the time it takes to hit the ground using the vertical component and then use that time to plug into see how far it goes in the horizontal component. Then, simply maximize that with respect to \theta. If you draw it out and do all that, you get:

untitledJPG.png


Vertical components:
Initial Velocity: v_0\sin{(\theta+\phi)}
Acceleration from Gravity: -g\cos{\phi}

Horizontal components:
Initial Velocity: v_0\cos{(\theta+\phi)}
Acceleration: +g\sin{\phi}

Time to hit ground:
0=v_0t\sin{(\theta+\phi)}-\frac{1}{2}gt^2\cos{\phi}\Rightarrow
t=\frac{2v_0\sin{\theta+\phi}}{g\cos{\phi}}

Now plug that t into:
v_0t\cos{(\theta+\phi)}+\frac{1}{2}gt^2\sin{\phi}

and maximize. Now from pure, unrelated experimentation, I also get \theta=\frac{90-\phi}{2} to maximize the distance.

Now my questions are

1) Is there anything wrong with my method? It's not a priority question but it would be useful
2) Is there a better way to do this? This solution is very messy and I was wondering if there was a more elegant, creative way to solve this. :)
3) Are both answers that I obtained above correct?

Thanks a lot

(I realize that another thread was made about this a year ago, but that never went anywhere, the problem was not solved, and the methods attempted was completely different, etc. so I thought it would be better simply to create a new topic. Merge if necessary: https://www.physicsforums.com/showthread.php?t=201484 )
 
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Welcome to PF!

Hi Anaxerzia ! Welcome to PF! :smile:
Anaxerzia said:
Pretend the ground is flat and that instead gravity is at an angle \phi-90), and that the ball is launch at angle \theta+\phi

What you are doing is using p and q coordinates (say), where p = x cosφ + y sinφ, q = y cosφ - x sinφ

Yes, that's a perfectly valid method! :smile:

(But "pretend the ground is flat" is not mathematical language! :wink:)
 
Or I guess you could have said to rotate by \phi , right? :-)

Does anyone have a more elegant solution? It seems to be lurking in there somewhere...
 

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