Range formula of a complicated projection

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You fire a ball with an initial speed $v_0$ at an angle $\phi$ above the surface of an incline, which is itself inclined at an angle $\theta$ above the horizontal (Fig. $3.37$).(a) Find the distance, measured along the incline, from the launch point to the point when the ball strikes the incline. (b) What angle $\phi$ gives the maximum range, measured along the incline ? Ignore air resistance.

I used the concept of projectile motion and used the formula$$R=\frac{u^2\cdot sin2\alpha}{g_0}$$
where $u=$initial velocity, $\alpha$=angle of projection and $g_0$=effective acceleration due to gravity

Here $u=v_0$, $\alpha=\phi$ and $g_0=g cos \theta$

Putting in the equation, $$R=\frac{v_0^2sin2\phi}{gcos \theta}$$

Is this formula correct ? If no, what is the correct formula?

Answer 1

It's very easy to get confused by these problems, especially as there are now two angles in play ($\theta$ and $\phi$).

Yet the solution is fairly straightforward.

Firstly, consider the angle of launch to be $\theta+\phi$.

Now compute a function, call it $y_p$, that describes the projectile's flight path as:

$$y_p=f(\theta+\phi, v_0, x)$$

You'll find this in Wikipedia.

Now define a second function, call it $y_s$, that defines the slope (a line) going through $(0,0)$ at angle $\theta$:

$$y_s=g(\theta, x)$$

The projectile hits the slope when:

$$y_p=y_s$$

Call that point $y_c$. To find the optimum angle $\phi$, compute:

$$\frac{\mathrm{d}y_c}{\mathrm{d}\phi}=0$$