Projectile on an Inclined Slope

A ball is thrown up a slope at angle $\alpha$ from the foot of the slope, with launch speed $v_0$ at angle $\theta$ from the horizontal. Where does it land back on the slope? This is the classic projectile-on-incline problem and it shows up in skiing, rocket launches from sloped pads, and any artillery in mountainous terrain.

The Physics

The ball follows the usual parabola from the launch point:

$x(t) = v_0\cos\theta\,t \qquad y(t) = v_0\sin\theta\,t - \tfrac{1}{2}\,g\,t^{2}$

The slope is the line $y = x\tan\alpha$. The ball lands when its trajectory intersects this line:

$v_0\sin\theta\,t - \tfrac{1}{2}\,g\,t^{2} = v_0\cos\theta\,t\tan\alpha$

Excluding the trivial $t = 0$ at launch and dividing by $t$:

$v_0\sin\theta - \tfrac{1}{2}\,g\,t = v_0\cos\theta\tan\alpha$

$t_{\text{land}} = \dfrac{2 v_0(\sin\theta - \cos\theta\tan\alpha)}{g}$

Step-by-Step Solution

Given:

• Launch speed: $v_0 = 15\;\text{m/s}$
• Launch angle (from horizontal): $\theta = 50°$
• Slope angle: $\alpha = 20°$
• Gravity: $g = 9.81\;\text{m/s}^{2}$

Find: The landing point on the slope, the flight time, and the distance traveled along the slope.

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Step 1 — Decompose the launch velocity.

$v_{0x} = 15\cos 50° = 15 \times 0.6428 \approx 9.642\;\text{m/s}$

$v_{0y} = 15\sin 50° = 15 \times 0.7660 \approx 11.491\;\text{m/s}$

Step 2 — Compute $\tan\alpha$ for the slope.

$\tan 20° \approx 0.3640$

Step 3 — Solve for the landing time using the formula above.

$t_{\text{land}} = \dfrac{2 v_0(\sin\theta - \cos\theta\tan\alpha)}{g}$

$= \dfrac{2(15)\bigl(\sin 50° - \cos 50° \cdot \tan 20°\bigr)}{9.81}$

$= \dfrac{30 \times (0.7660 - 0.6428 \times 0.3640)}{9.81}$

$= \dfrac{30 \times (0.7660 - 0.2340)}{9.81}$

$= \dfrac{30 \times 0.5320}{9.81} = \dfrac{15.961}{9.81} \approx 1.627\;\text{s}$

Step 4 — Find the landing coordinates $(x_{\text{land}}, y_{\text{land}})$.

$x_{\text{land}} = v_{0x}\,t_{\text{land}} = 9.642 \times 1.627 \approx 15.69\;\text{m}$

$y_{\text{land}} = x_{\text{land}}\tan\alpha = 15.69 \times 0.3640 \approx 5.711\;\text{m}$

(Check: this should equal $v_{0y}\,t - \tfrac{1}{2}gt^{2} = 11.491 \times 1.627 - 4.905 \times 1.627^{2} = 18.696 - 12.985 \approx 5.711\;\text{m}$. ✓)

Step 5 — Find the distance along the slope (the actual "range").

The landing point is at $(15.69, 5.711)$ from the launch origin. The straight-line distance is:

$L = \sqrt{x_{\text{land}}^{2} + y_{\text{land}}^{2}} = \sqrt{(15.69)^{2} + (5.711)^{2}}$

$L = \sqrt{246.18 + 32.62} = \sqrt{278.80} \approx 16.70\;\text{m}$

Step 6 — (Bonus) Compute the optimal launch angle for this slope.

A famous result: to maximize the slope distance $L$ for a given $v_0$ and $\alpha$, the optimal launch angle (measured from horizontal) is:

$\theta_{\text{opt}} = 45° + \dfrac{\alpha}{2} = 45° + 10° = 55°$

So at $\theta = 50°$ we're slightly below optimal — bumping to 55° would give us a few extra meters along the slope.

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Answer: The ball lands on the slope at the point $(\approx 15.69, \approx 5.71)$ meters from the launch point, after a flight time of $\approx 1.63\;\text{s}$. The distance traveled along the slope is $L \approx 16.70\;\text{m}$. The optimal launch angle for this 20° slope would be 55° (giving slightly more range).

Try It

• Tilt the slope with the slope-angle slider — see how the landing point shifts. Negative $\alpha$ gives a downhill launch.
• Set the slope to 0° to recover the classic ground-to-ground problem ($\theta_{\text{opt}} = 45°$).
• Try a launch angle equal to the slope angle ($\theta = \alpha$) — the trajectory is tangent to the slope and the ball never lands cleanly.
• The HUD shows the optimal angle for the current slope — try matching it.