Maximum range for projectile thrown from a mountain [closed]

I have doubts with the following problem:

A projectile is fired from the top of a mountain that has a downward slope with angle $\phi$ from the horizontal. The initial velocity of the projectile is $v_{0}$ and has a angle $\theta$. Show that the range $R$ (maximum horizontal distance traveled) is related to $v_{0}$, $\theta$ and $\phi$ for:

$$R=\frac{v_{0}^{2}}{g}(\sin2\theta+\tan\phi(1+\cos2\theta))$$

To accomplish this, I began by analyzing the forces acting on the projectile. Then, using Newton's second law, I found the equations of motion:

$$x(t)=(v_{0}\cos\theta) t$$

$$y(t)=(v_{0}\sin\theta) t - \frac{1}{2}gt^{2}$$

As we reach the maximum, then I get the time $t_{R}$ it takes the projectile to touch ground:

$$y(t_{R})=-d=(v_{0}\sin\theta) t_{R} - \frac{1}{2}gt_{R}^{2}$$

Using the general formula and taking into account the positive sign:

$$t_{R}=\frac{v_{0}\sin\theta+\sqrt{v^2_{0}\sin^2\theta+2gd}}{g}$$

Substituting into the equation $x(t)$:

$$x(t_{R})=R=(v_{0}\cos\theta) \frac{v_{0}\sin\theta+\sqrt{v^2_{0}\sin^2\theta+2gd}}{g}$$

In this equation, I should get $R$, but I failed to remove the square root and I also have that $d=R\tan\phi$.

My questions are: Did I make a mistake in the procedure? Is there a simpler way to solve the problem? Or someone could give me a Hint to get $R$, starting from the last equation. By dividing the path into two parts, came to the same equation.

closed as off-topic by John Rennie, jinawee, Kyle Kanos, BebopButUnsteady, Brandon EnrightFeb 25 '14 at 21:42

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Hint : Why don't you change your co-ordinate axes? x-axis to down the incline and y-axis normally upwards. Hence, you acceleration will have 2 components :

$a_x = g \sin\phi$

$a_y= - g \cos\phi$

You can now process as usual and note that your when projectile falls on incline for first time, its y-coordinate is $0$.

x-coordinate is length $on$ the incline.

These types of questions are generally easier to solve using this approach. This also makes you realize how much you are accustomed to your "general" coordinate system and never tend to change (or even rotate) it.

I do not think you made any mistakes. Try rewriting your final equation with $d=R\tan\phi$ in the form $\sqrt{...}= \mbox{the rest}$. Then solve it for $R$.

Cheers, Frédéric