# Projectile thrown up an incline, max distance?

1. Jun 16, 2009

### sabinscabin

1. The problem statement, all variables and given/known data

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
$$R_{max} = \frac{ V^2 }{ g (1 + \sin{\phi} )}$$

2. Relevant equations

F = ma

3. 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?

2. Jun 16, 2009

### glueball8

3. Jun 16, 2009

### sabinscabin

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.

4. Jun 16, 2009

### glueball8

5. Jun 16, 2009

### ideasrule

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.

6. Jun 16, 2009

### 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$$\varphi$$)