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A ball is thrown with initial speed $v_0$ up an inclined plane. The plane is inclined at an angle $\phi$ above the horizontal, and the ball's initial velocity is at an angle $\theta$ above the plane. Choose axes with $x$ measured up the slope, $y$ normal to the slope, 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 $R = \frac{2v_0^2\sin\theta\cos(\theta + \phi)}{g\cos^2\phi}$ from its launch point. Show that for given $v_0$ and $\phi$, the maximum possible range up the inclined plane is $R_{\max} = \frac{v_0^2}{g(1 + \sin\phi)}$.

Here is my drawing:

Newton's $2^{\text{nd}}$ is $\vec{F} = m\vec{a}$ where we can write $\vec{F} = F_x\vec{x} + F_y\vec{y}$.

\begin{alignat*}{3}

F_x & = & m\ddot{x}\\

F_y & = & m\ddot{y}

\end{alignat*}

Now what should I do? Also, am I correct so far?