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## Homework Statement

A ball is thrown with initial speed [tex] V [/tex] up an inclined plane. The plane is inclined at an angle [tex]\phi[/tex] above the horizontal, and the ball's velocity is at an angle [tex]\theta[/tex] above the plane.

Show that the ball lands a distance [tex]R = \frac{ 2V^2 \sin{\theta} \cos{ ( \theta + \phi} ) } { g \cos^2{\phi} }[/tex] from its launch point. Show that for a given [tex]V[/tex] and

[tex]\phi[/tex], the maximum possible range up the inclined plane is

[tex] R_{max} = \frac{ V^2 }{ g (1 + \sin{\phi} )} [/tex]

## Homework Equations

F = ma

## The Attempt at a Solution

I calculated the distance traveled up the incline fine. However, I'm having trouble proving the second part. I'm guessing I'm supposed to maximize R with respect to theta, so from the equation above we have:

[tex]\frac{d} {d \theta} \sin{\theta} \cos{(\theta + \phi)} = 0[/tex]

[tex]\cos{( 2 \theta + \phi )} = 0[/tex]

[tex]\theta = \frac{n \pi}{4} - \frac{\phi}{2} [/tex], with n = odd integer. Now plug this back into the original equation for R and I get

[tex]R_{max} = \frac{ 4V^2 \tan{\phi} }{g} \cos{ 2\phi} [/tex]

I think I'm approaching this problem wrong. Can anyone give me a simpler way?