Projectile on Inclined Plane — Range Up the Plane

The situation

A plane is inclined at angle $\beta$ to the horizontal. A body is projected from the bottom of the plane with speed $u$ at angle $\alpha$ above the horizontal. It lands somewhere on the inclined surface above. The distance along the slope from launch to landing is $R$.

Setting up axes along and perpendicular to the plane

The key to this problem is choosing axes wisely. Instead of the standard horizontal/vertical axes, we use:

• $x$-axis: along the inclined plane (up the slope = positive)
• $y$-axis: perpendicular to the inclined plane (away from surface = positive)

The body lands back on the plane, so the perpendicular displacement at landing is zero: $y = 0$.

Resolving initial velocity and gravity

Initial velocity $u$ is at angle $\alpha$ to the horizontal, which means it is at angle $(\alpha - \beta)$ to the inclined plane.

Components along the new axes:

• Along the plane: $u_x = u\cos(\alpha - \beta)$
• Perpendicular to plane: $u_y = u\sin(\alpha - \beta)$

Gravity $g$ acts vertically downward. Resolving along the new axes:

• Along the plane (down the slope): $g\sin\beta$
• Perpendicular to plane (into the surface): $g\cos\beta$

With our sign convention (up slope and away from surface positive):

• Acceleration along plane: $a_x = -g\sin\beta$
• Acceleration perpendicular to plane: $a_y = -g\cos\beta$

Finding time of flight

The body lands when perpendicular displacement $= 0$:

$y = u_y T + \frac{1}{2}a_y T^2 = 0$

$u\sin(\alpha-\beta) \cdot T - \frac{1}{2}g\cos\beta \cdot T^2 = 0$

$T\left[u\sin(\alpha-\beta) - \frac{1}{2}g\cos\beta \cdot T\right] = 0$

Taking $T \neq 0$:

$T = \frac{2u\sin(\alpha-\beta)}{g\cos\beta}$

Finding the range

Apply the displacement equation along the plane:

$R = u_x T + \frac{1}{2}a_x T^2$

$R = u\cos(\alpha-\beta) \cdot T - \frac{1}{2}g\sin\beta \cdot T^2$

$R = T\left[u\cos(\alpha-\beta) - \frac{1}{2}g\sin\beta \cdot T\right]$

Substitute $T = \frac{2u\sin(\alpha-\beta)}{g\cos\beta}$:

$R = \frac{2u\sin(\alpha-\beta)}{g\cos\beta}\left[u\cos(\alpha-\beta) - \frac{1}{2}g\sin\beta \cdot \frac{2u\sin(\alpha-\beta)}{g\cos\beta}\right]$

$R = \frac{2u\sin(\alpha-\beta)}{g\cos\beta}\left[u\cos(\alpha-\beta) - \frac{u\sin\beta\sin(\alpha-\beta)}{\cos\beta}\right]$

$R = \frac{2u^2\sin(\alpha-\beta)}{g\cos\beta}\left[\cos(\alpha-\beta) - \frac{\sin\beta\sin(\alpha-\beta)}{\cos\beta}\right]$

$R = \frac{2u^2\sin(\alpha-\beta)}{g\cos^2\beta}\left[\cos(\alpha-\beta)\cos\beta - \sin\beta\sin(\alpha-\beta)\right]$

The bracket is $\cos[(\alpha-\beta)+\beta] = \cos\alpha$:

$\boxed{R = \frac{2u^2\sin(\alpha-\beta)\cos\alpha}{g\cos^2\beta}}$

Condition for the formula to apply

This formula is valid only when the body lands on the inclined surface above the launch point, i.e. $\alpha > \beta$ (the projection angle must be greater than the slope angle).

If $\alpha < \beta$, the body is projected into the slope and does not become a projectile on the surface.

If $\alpha = \beta$, the body is projected along the surface ($R = 0$).