Projectile thrown up an incline, max distance?

[sabinscabin]

Homework Statement

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

Show that the ball lands a distance R = \frac{ 2V^2 \sin{\theta} \cos{ ( \theta + \phi} ) } { g \cos^2{\phi} } from its launch point. Show that for a given V and
\phi, the maximum possible range up the inclined plane is
R_{max} = \frac{ V^2 }{ g (1 + \sin{\phi} )}

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:

\frac{d} {d \theta} \sin{\theta} \cos{(\theta + \phi)} = 0

\cos{( 2 \theta + \phi )} = 0

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

R_{max} = \frac{ 4V^2 \tan{\phi} }{g} \cos{ 2\phi}

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

 

[glueball8]

https://www.physicsforums.com/showthread.php?t=319978

 

[sabinscabin]

https://www.physicsforums.com/showthread.php?t=319978

if you look at the last post in the thread you gave me, I already got that result. It's just that the R_max I get does not match the R max that's given in the initial problem.

 

[glueball8]

if you look at the last post in the thread you gave me, I already got that result. It's just that the R_max I get does not match the R max that's given in the initial problem.

Ops my bad.

 

[ideasrule]

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

R_{max} = \frac{ 4V^2 \tan{\phi} }{g} \cos{ 2\phi}

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

First, n must be 1; if it's 3 or higher, than n*pi/4 - Φ/2 would be greater than 90 degrees, and that's impossible given the geometry of the situation. So we know that the optimum angle is pi/4-Φ/2. Substituting it into the equation for R gives:

R=sin(pi/4-Φ/2)cos(pi/4+Φ/2)/cos^2 (Φ)

I've ignored the constants. Since cos(a)=sin(pi/2-a):

R=sin(pi/4-Φ/2)sin(pi/4-Φ/2)
cos(2a)=1-2sin^2 (a), so

R=(1-cos(pi/2-Φ))/2
=(1-sin(Φ))/2

You seem to have decent math skills, so getting from here to the answer should be easy.

 

[SimonZ]

you have got \theta = \pi/4 - \varphi/2
so 2sin\theta cos(\theta + \varphi) = 2sin(\pi/4 - \varphi/2) sin(\pi/4 - \varphi/2) =
2sin²(\pi/4 - \varphi/2) = [cos(\varphi/2) - sin(\varphi/2)]²

[cos(\varphi/2) - sin(\varphi/2)]²/cos²\varphi
= {[cos(\varphi/2) - sin(\varphi/2)]/cos\varphi}²
= {[cos(\varphi/2) - sin(\varphi/2)]/[cos²(\varphi/2) - sin²(\varphi/2)]}²
= {1/[cos(\varphi/2) + sin(\varphi/2)]}²
= 1/(1 + sin\varphi)