1. ## classical mechanics problem

A ball is thrown with initial speed $\displaystyle v_0$ up an inclined plane. the plane is at an angle $\displaystyle \phi$ and the ball's initial velocity is at angle $\displaystyle \theta$. Choose an axis with x measured up the slope, y normal, and z across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance

$\displaystyle R=\dfrac{2v_0^2\sin{\theta}\cos(\theta+\phi)}{\cos ^2(\phi)}$

from it's launch point.

Show that for a given $\displaystyle v_0$ and $\displaystyle \phi$ that the max range is $\displaystyle R_{max}=\dfrac{v_0^2}{g(1+\sin\theta)}$

2. Um... have you tried anything?

3. I've tried many things but none of them get me even close to getting that result. I know that the force in the x direction is $\displaystyle mg\sin\theta$ and so the displacement formula has this subed for g without the for the displacement in the x direction. The displacement in the x direction is $\displaystyle v_0\cos\theta t$ and the y is normal so there's no displacement in the y.

4. Ok I'm coming back to this now.

I took a look at the derivation for this problem in two dimensions for a projectile and tried to model my approach after it. This is how far I got.

$\displaystyle 0=x(t)= v_0 \sin(\phi)t-\frac{1}{2}g \cos(\theta)t^2$

then from this we can get

$\displaystyle t=\dfrac{2v_0\sin(\phi)}{g\cos(\theta)}$

Then as in the 2d simple case I multiplied $\displaystyle t v_0$ to obtain $\displaystyle R$

but this doesn't give the result I was looking for.

Can anyone help me with this?

5. Originally Posted by magus
A ball is thrown with initial speed $\displaystyle v_0$ up an inclined plane. the plane is at an angle $\displaystyle \phi$ and the ball's initial velocity is at angle $\displaystyle \theta$. Choose an axis with x measured up the slope, y normal, and z across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance

$\displaystyle R=\dfrac{2v_0^2\sin{\theta}\cos(\theta+\phi)}{g\co s^2(\phi)}$

note the correction in your formula for R from your original post.
$\displaystyle \displaystyle \Delta x = v_0\cos{\theta} \cdot t - \frac{1}{2}g\sin{\phi} \cdot t^2$

$\displaystyle \displaystyle \Delta y = v_0\sin{\theta} \cdot t - \frac{1}{2}g\cos{\phi} \cdot t^2$

since $\displaystyle \Delta y = 0$ ...

$\displaystyle \displaystyle t = \frac{2v_0\sin{\theta}}{g\cos{\phi}}$

substituting for $\displaystyle t$ in the $\displaystyle \Delta x$ equation ...

$\displaystyle \displaystyle \Delta x = \frac{2v_0^2 \sin{\theta} \cos{\theta}}{g\cos{\phi}} - \frac{2v_0^2 \sin^2{\theta} \sin{\phi}}{g\cos^2{\phi}}$

common denominator ...

$\displaystyle \displaystyle \Delta x = \frac{2v_0^2 \sin{\theta} \cos{\theta}\cos{\phi}}{g\cos^2{\phi}} - \frac{2v_0^2 \sin^2{\theta} \sin{\phi}}{g\cos^2{\phi}}$

combine and factor ...

$\displaystyle \displaystyle \Delta x = \frac{2v_0^2 \sin{\theta}(\cos{\theta}\cos{\phi} - \sin{\theta}\sin{\phi})}{g\cos^2{\phi}}$

finally, note that ...

$\displaystyle \cos{\theta}\cos{\phi} - \sin{\theta}\sin{\phi} = \cos(\theta+\phi)$

... and you're there.

6. I think there is a 'g' missing in your first part.

Make a sketch.

(I'll be using v instead of vo to make it easier to type)

The velocity of the ball up the plane is $\displaystyle v \cos\theta$
That perpendicular to the plane is $\displaystyle v \sin\theta$

The acceleration is down the plane and is given by $\displaystyle g\ sin\phi$
And that perpendicular to the plane is given by $\displaystyle g \cos\phi$

From this, the distance perpendicular to the plane where the particle lands is 0 and is given by:

$\displaystyle s = ut + \dfrac12 at^2$

This:

$\displaystyle 0 = v\sin\theta t - \dfrac12 g\cos\phi t^2$

For the distance along the plane, we have:

$\displaystyle s = v\cos\theta t - \dfrac12 g\sin\phi t^2$

Can you complete the first part now?

EDIT: Didn't see you replied Skeeter

7. For the second part now. (there is also a mistake, see below)

$\displaystyle R = \dfrac{2v^2\sin\theta\cos(\theta+\phi)}{g\cos^2\ph i}$

Use the identity: $\displaystyle 2\sin A\cos B = \sin(A+B) + \sin(A-B)$

This gives:

$\displaystyle R = \dfrac{v^2(\sin(2\theta + \phi) + \sin(-\phi))}{g\cos^2(\phi)}$

Simplify:

$\displaystyle R = \dfrac{v^2(\sin(2\theta + \phi) - \sin(\phi))}{g(1+\sin\phi)(1 - \sin\phi)}$

At the maximum range, $\displaystyle \sin(2\theta + \phi) = 1$ since only theta can vary.

Hence we get:

$\displaystyle R = \dfrac{v^2(1 - \sin(\phi))}{g(1+\sin\phi)(1 - \sin\phi)}$

Something cancels out, giving:

$\displaystyle R_{max} = \dfrac{v^2}{g(1+\sin\phi)}$

From this, you can even find the relation between theta and phi for this value of range

8. Thanks. The $\displaystyle \Delta y$ is the other equation I really needed I guess.