Projectile on an Inclined Plane

When a projectile is launched along (or against) an inclined plane, the analysis is cleanest in rotated axes — one along the incline, one perpendicular.

Concept

Let the incline make an angle $\alpha$ with the horizontal. Project an object with speed $u$ at angle $\theta$ measured from the incline.

Rotated Frame

Take axes along the incline ($x'$) and perpendicular ($y'$). Gravity has components:

• Along incline: $g\sin\alpha$ (down the slope).
• Perpendicular to incline: $g\cos\alpha$ (into the slope).

Initial velocity components:

• $u_{x'} = u\cos\theta$.
• $u_{y'} = u\sin\theta$.

Equations of Motion

$x'(t) = u\cos\theta\,t - \tfrac{1}{2}g\sin\alpha\,t^2$ $y'(t) = u\sin\theta\,t - \tfrac{1}{2}g\cos\alpha\,t^2$

Time of Flight

Particle returns to incline when $y' = 0$: $T = \frac{2u\sin\theta}{g\cos\alpha}.$

Range Up the Incline

$R = u\cos\theta \cdot T - \tfrac{1}{2}g\sin\alpha \cdot T^2.$

After simplification: $\boxed{R = \frac{2u^2\sin\theta\cos(\theta + \alpha)}{g\cos^2\alpha}.}$

Maximum Range

To maximize $R$ over $\theta$, set $dR/d\theta = 0$. This yields: $\theta_{\max} = \frac{\pi}{4} - \frac{\alpha}{2} = 45° - \frac{\alpha}{2}$

with maximum range $R_{\max} = \frac{u^2}{g(1 + \sin\alpha)}.$

Range Down the Incline

By symmetry (replace $\alpha \to -\alpha$): $R_{\text{down}} = \frac{2u^2\sin\theta\cos(\theta - \alpha)}{g\cos^2\alpha},$ with maximum at $\theta = 45° + \alpha/2$ and $R_{\text{down, max}} = \frac{u^2}{g(1 - \sin\alpha)}.$

So shooting down a slope reaches further than shooting up — as expected.

Worked Example

Q: A ball is projected up a 30° incline at speed $20\,\text{m/s}$ at angle $30°$ measured from the incline. Take $g = 10\,\text{m/s}^2$. Find the range up the incline.

Solution: $\theta = 30°$, $\alpha = 30°$, $u = 20$.

Time of flight: $T = (2 \cdot 20 \cdot \sin 30°)/(10 \cos 30°) = 20/(10 \cdot \sqrt{3}/2) = 4/\sqrt{3} \approx 2.31$ s.

Range: $R = \frac{2(400)\sin 30°\cos 60°}{10\cos^2 30°} = \frac{2(400)(0.5)(0.5)}{10(3/4)} = \frac{200}{7.5} \approx 26.67\,\text{m}.$

Maximum range (over angles) for this incline: $R_{\max} = \frac{400}{10(1 + 0.5)} = \frac{400}{15} \approx 26.67\,\text{m}.$

So the chosen angle here is also the optimal one for this incline ($\theta_{\max} = 45° - 15° = 30°$ ✓).

Common Confusions

• "Use horizontal/vertical axes." — You can, but it's much messier; rotated axes simplify enormously.
• "$\theta$ is from horizontal." — In the formula above, $\theta$ is measured from the incline surface.
• "$\theta_{\max} = 45°$ always." — Only for flat ground ($\alpha = 0$). For an upward incline, $\theta_{\max} < 45°$ (relative to incline).

Key Takeaways

• Use axes along and perpendicular to the incline.
• Range up the incline: $R = 2u^2\sin\theta\cos(\theta + \alpha)/(g\cos^2\alpha)$.
• Optimal angle (up): $45° - \alpha/2$.
• Max range up: $u^2/[g(1 + \sin\alpha)]$.
• Down the incline reaches farther than up: $R_{\text{down,max}} = u^2/[g(1 - \sin\alpha)]$.