How to Derive the Range of a Projectile on an Inclined Plane?

[Redwaves]

Homework Statement

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.

Relevant Equations

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

$V_x = V_0 cos \theta$
$x = V_0 cos \theta t$

$V_y = V_0 cos \theta$
$y = V_0 cos \theta t$

$F_x = m\ddot{x}$
$-mgsin \phi = m\ddot{x}$
$\dot{x} = -gtsin\phi + V_x$
$x = -\frac{1}{2} gt^2 sin \phi + V_x t$
$x = -\frac{1}{2} gt^2 sin \phi + v_0 cos\theta t$

$F_y = m\ddot{y}$
$-mgcos \phi = m\ddot{y}$
$\dot{y} = -gtcos\phi + V_y$
$y = -\frac{1}{2} gt^2 cos \phi + V_y t$
$y = -\frac{1}{2} gt^2 cos \phi + v_0 sin \theta t$

I don't see where $R$ comes from.

 

[haruspex]

You have some transcription errors in some equations (and what are x0, y0?), but you final equations are ok.
What is the value of y when it lands?

 

[Redwaves]

You have some transcription errors in some equations (and what are x0, y0?), but you final equations are ok.
What is the value of y when it lands?

y = 0, since the ball is on the inclined plane?

 

[haruspex]

y = 0, since the ball is on the inclined plane?

Yes. So what do you deduce for x at that point?

 

[Redwaves]

Yes. So what do you deduce for x at that point?

I have to find t from the second equation then replace t in the x equation. Is it correct?

 

[haruspex]

I have to find t from the second equation then replace t in the x equation. Is it correct?

Yes.

 

[Redwaves]

I found it, thanks

 

[Fred Wright]

I suggest you ignore the bit about measuring x along the slope and y normal to the slope. In that case we have,
$$
v_{0x}=|v_0| \cos(\theta +\phi)
$$
$$
v_{0y}=|v_0| \sin(\theta +\phi)

$$
Neglecting wind resistance, the only force is in the y direction.
$$
ma_y=-mg
$$
$$
\frac{dv_y}{dt}=-g
$$
$$
y=v_{0y}t-\frac{1}{2}gt^2
$$
From $\frac{dv_{0x}}{dt}=0$, the velocity in the x direction is constant, i.e. $v_x=v_{0x}$.
$$
\frac{dx}{dt}=v_{0x}
$$
$$
t=\frac{x}{v_{0x}}
$$
Plugging the expression for $t$ into our equation for y we get
$$
y=\frac{v_{0y}}{v_{0x}}x-\frac{1}{2}g(\frac{x}{v_{0x}})^2
$$
The trajectory intersects the plane at
$$
y=R\sin(\phi)
$$
$$
x=R\cos(\phi)
$$
We thus have
$$
R\sin(\phi)=R\cos(\phi)\frac{v_{0y}}{v_{0x}}-\frac{1}{2}g\frac{R^2\cos^2(\phi)}{v_{0x}^2}
$$
Solving for $R$,
$$
R=\frac{2|v|^2\cos(\theta + \phi)}{g\cos^2(\phi)}(\sin(\theta + \phi)\cos(\phi)-\cos(\theta + \phi)\sin(\phi))
$$
From trig addition angle formulas,
$$
\sin(\theta + \phi)=\sin(\theta)\cos(\phi)+\sin(\phi)\cos(\theta)
$$
$$
\cos(\theta + \phi)=\cos(\theta)\cos(\phi)-\sin(\phi)\sin(\theta)
$$
$$
R=\frac{2|v|^2\cos(\theta + \phi)}{g\cos^2(\phi)}(\sin(\theta)\cos^2(\phi)+\sin(\phi)\cos(\phi)\cos(\theta) -\sin(\phi)\cos(\phi)\cos(\theta) +\sin^2(\phi)\sin(\theta))
$$
$$
R=\frac{2|v|^2\cos(\theta + \phi)\sin(\theta)}{g\cos^2(\phi)}
$$

 

[Delta2]

Jesus @Fred Wright , I haven't checked your solution in detail but it seems to me that you reveal the whole solution to the OP. Here in physics forums we are not allowed to do that. My only hope is that the OP is obliged to use x and y-axis as mentioned in the OP

 

[haruspex]

Jesus @Fred Wright , I haven't checked your solution in detail but it seems to me that you reveal the whole solution to the OP. Here in physics forums we are not allowed to do that. My only hope is that the OP is obliged to use x and y-axis as mentioned in the OP

My reading of post #7 is that, fortunately, @Redwaves solved it already using the prescribed method. And it doesn't seem to me that the method in post #8 is much simpler, if at all.