Projectile on Inclined Plane — Range Down the Plane

The situation

A body is projected from the top edge of an inclined plane (angle $\beta$) with speed $u$ at angle $\alpha$ above the horizontal. The body lands on the inclined surface below the launch point. The distance along the slope is $R$.

How this differs from projection up the plane

In the up-the-plane case, the body is projected toward the higher part of the slope. Here, the body is projected outward and lands on the lower part.

The setup is identical — we use axes along and perpendicular to the plane — but now the component of initial velocity along the plane is directed down the slope, and at landing the body is below the launch point.

Setting up axes

Same as before:

• $x$-axis: along the plane, but now taking down the slope as positive
• $y$-axis: perpendicular to the plane, away from surface = positive

The body is projected at angle $\alpha$ above horizontal. Since down the slope is now positive $x$, the angle between the initial velocity and the slope (measured toward the positive $x$ direction) is $(\alpha + \beta)$.

Components of initial velocity:

• Along the plane (down): $u_x = u\cos(\alpha+\beta)$
• Perpendicular (away): $u_y = u\sin(\alpha+\beta)$

Components of gravity along the new axes (gravity has a component down the slope and into the surface):

• Along the plane (down = positive): $a_x = g\sin\beta$
• Perpendicular (into surface = negative): $a_y = -g\cos\beta$

Finding time of flight

Body lands when perpendicular displacement $= 0$:

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

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

Finding range

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

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

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

Substitute $T$:

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

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

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

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

Comparison with up-the-plane formula

Since $\sin(\alpha+\beta) > \sin(\alpha-\beta)$ for $\beta > 0$, the range down the plane is always greater than up the plane for the same $u$ and $\alpha$. This makes physical sense — gravity assists the motion down the slope.