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Concept 08 / 10Kinematics

Projectile Motion on an Inclined Plane

Solving slope-landing projectiles by rotating the axes so gravity splits into g sin beta along the slope and g cos beta normal to it, giving time of flight, up-slope and down-slope range, and optimal launch angles.

Figure 1
When a projectile lands on a sloping surface, the ordinary ground-level formulae fail, because the landing condition is no longer y=0y = 0. The efficient trick is to rotate the axes: take x′x' along the incline (positive up the slope) and y′y' perpendicular to it. In these axes the landing condition becomes the simple y′=0y' = 0, at the price that gravity now has components along both axes. If the incline makes angle β\beta with the horizontal, then resolving g⃗\vec{g} gives ax′=−gsin⁡βa_{x'} = -g\sin\beta and ay′=−gcos⁡βa_{y'} = -g\cos\beta. The essential difference from ordinary projectile motion is that the 'horizontal' acceleration is no longer zero, as the figure above indicates. Let the projectile be launched with speed uu at angle α\alpha measured from the incline surface (so its angle with the true horizontal is α+β\alpha + \beta for an upward slope). Then ux′=ucos⁡αu_{x'} = u\cos\alpha and uy′=usin⁡αu_{y'} = u\sin\alpha. The perpendicular motion is ordinary free fall with effective gravity gcos⁡βg\cos\beta, so the time of flight follows from 0=usin⁡α T−12gcos⁡β T20 = u\sin\alpha\,T - \frac{1}{2}g\cos\beta\,T^{2}, giving T=2usin⁡αgcos⁡βT = \frac{2u\sin\alpha}{g\cos\beta}. Notice that TT does not involve sin⁡β\sin\beta at all; the along-slope acceleration cannot affect when the body returns to the surface. The range along the incline is then R=ucos⁡α T−12gsin⁡β T2R = u\cos\alpha\,T - \frac{1}{2}g\sin\beta\,T^{2}. Substituting TT and simplifying gives R=2u2sin⁡αcos⁡(α+β)gcos⁡2βR = \frac{2u^{2}\sin\alpha\cos(\alpha+\beta)}{g\cos^{2}\beta}. For a projectile fired down the slope, every occurrence of β\beta changes sign, so Rdown=2u2sin⁡αcos⁡(α−β)gcos⁡2βR_{down} = \frac{2u^{2}\sin\alpha\cos(\alpha-\beta)}{g\cos^{2}\beta}, while TT is unchanged in form. To maximise the up-slope range, use 2sin⁡αcos⁡(α+β)=sin⁡(2α+β)−sin⁡β2\sin\alpha\cos(\alpha+\beta) = \sin(2\alpha+\beta) - \sin\beta, so R=u2[sin⁡(2α+β)−sin⁡β]gcos⁡2βR = \frac{u^{2}\left[\sin(2\alpha+\beta)-\sin\beta\right]}{g\cos^{2}\beta}. Only the first term depends on α\alpha, and it is greatest when 2α+β=90∘2\alpha+\beta = 90^{\circ}, that is α=π4−β2\alpha = \frac{\pi}{4} - \frac{\beta}{2}: the optimal launch bisects the angle between the incline and the vertical, exactly as 45∘45^{\circ} bisects the angle between the horizontal ground and the vertical. Substituting gives Rmax=u2(1−sin⁡β)gcos⁡2β=u2(1−sin⁡β)g(1−sin⁡2β)=u2g(1+sin⁡β)R_{max} = \frac{u^{2}(1-\sin\beta)}{g\cos^{2}\beta} = \frac{u^{2}(1-\sin\beta)}{g(1-\sin^{2}\beta)} = \frac{u^{2}}{g(1+\sin\beta)}. For the down-slope case, α=π4+β2\alpha = \frac{\pi}{4} + \frac{\beta}{2} and Rmax=u2g(1−sin⁡β)R_{max} = \frac{u^{2}}{g(1-\sin\beta)}. Both reduce to u2g\frac{u^{2}}{g} at β=0\beta = 0, a good check, and they show that throwing downhill beats throwing uphill, as intuition demands. A recurring Advanced question is the condition for the projectile to strike the incline perpendicularly. Striking perpendicular to the surface means the along-slope velocity component vanishes at landing: ucos⁡α−gsin⁡β T=0u\cos\alpha - g\sin\beta\,T = 0. Substituting T=2usin⁡αgcos⁡βT = \frac{2u\sin\alpha}{g\cos\beta} gives ucos⁡α=2ugsin⁡αsin⁡βgcos⁡βu\cos\alpha = \frac{2ug\sin\alpha\sin\beta}{g\cos\beta}, so cot⁡α=2tan⁡β\cot\alpha = 2\tan\beta. Another common variant asks for the projectile to land back at the launch point, which requires the along-slope displacement to be zero, ucos⁡α T=12gsin⁡β T2u\cos\alpha\,T = \frac{1}{2}g\sin\beta\,T^{2}, giving cot⁡α=tan⁡β\cot\alpha = \tan\beta, i.e. α=90∘−β\alpha = 90^{\circ}-\beta, so that α+β=90∘\alpha+\beta = 90^{\circ}. Since α+β\alpha+\beta is the angle with the true horizontal, this says the throw is vertically upward, which is obvious once seen: only a vertical throw comes straight back down to where it started. Key Takeaways: - Rotate axes: x′x' along the slope, y′y' normal to it; ax′=−gsin⁡βa_{x'} = -g\sin\beta, ay′=−gcos⁡βa_{y'} = -g\cos\beta. - With α\alpha measured from the incline: T=2usin⁡αgcos⁡βT = \frac{2u\sin\alpha}{g\cos\beta} (same for up- and down-slope). - Up the slope: R=2u2sin⁡αcos⁡(α+β)gcos⁡2βR = \frac{2u^{2}\sin\alpha\cos(\alpha+\beta)}{g\cos^{2}\beta}, maximum at α=π4−β2\alpha = \frac{\pi}{4}-\frac{\beta}{2}, Rmax=u2g(1+sin⁡β)R_{max} = \frac{u^{2}}{g(1+\sin\beta)}. - Down the slope: R=2u2sin⁡αcos⁡(α−β)gcos⁡2βR = \frac{2u^{2}\sin\alpha\cos(\alpha-\beta)}{g\cos^{2}\beta}, maximum at α=π4+β2\alpha = \frac{\pi}{4}+\frac{\beta}{2}, Rmax=u2g(1−sin⁡β)R_{max} = \frac{u^{2}}{g(1-\sin\beta)}. - Perpendicular impact on the incline: cot⁡α=2tan⁡β\cot\alpha = 2\tan\beta. Return to the launch point: cot⁡α=tan⁡β\cot\alpha = \tan\beta, i.e. α+β=90∘\alpha+\beta = 90^{\circ}, a vertical throw. - Maximum distance from the incline (not the maximum height above the ground) is u2sin⁡2α2gcos⁡β\frac{u^{2}\sin^{2}\alpha}{2g\cos\beta}. - Pitfall: mixing up α\alpha (from the incline) with the angle from the horizontal, which is α+β\alpha+\beta. - Pitfall: setting ax′=0a_{x'} = 0 out of habit; along the slope the acceleration is gsin⁡βg\sin\beta, not zero. Brain-Teaser 1: A particle is projected with u=10u = 10 m/s at 30∘30^{\circ} to a plane inclined at 30∘30^{\circ} to the horizontal, up the slope. Take g=10g = 10 m/s2^{2}. Find the time of flight and the range along the incline, and check whether this launch angle happens to be the optimal one. Solution: Here α=30∘\alpha = 30^{\circ} and β=30∘\beta = 30^{\circ}. Time of flight T=2usin⁡αgcos⁡β=2(10)(0.5)10(32)=1053=23≈1.155T = \frac{2u\sin\alpha}{g\cos\beta} = \frac{2(10)(0.5)}{10\left(\frac{\sqrt{3}}{2}\right)} = \frac{10}{5\sqrt{3}} = \frac{2}{\sqrt{3}} \approx 1.155 s. Range R=2u2sin⁡αcos⁡(α+β)gcos⁡2β=2(100)(0.5)cos⁡60∘10(34)=100×0.57.5=507.5=203≈6.67R = \frac{2u^{2}\sin\alpha\cos(\alpha+\beta)}{g\cos^{2}\beta} = \frac{2(100)(0.5)\cos 60^{\circ}}{10\left(\frac{3}{4}\right)} = \frac{100 \times 0.5}{7.5} = \frac{50}{7.5} = \frac{20}{3} \approx 6.67 m. Optimality: the best angle is α=45∘−β2=45∘−15∘=30∘\alpha = 45^{\circ} - \frac{\beta}{2} = 45^{\circ} - 15^{\circ} = 30^{\circ}, which is exactly the given angle, so this is the maximum range. Confirm with the closed form: Rmax=u2g(1+sin⁡β)=10010(1.5)=203R_{max} = \frac{u^{2}}{g(1+\sin\beta)} = \frac{100}{10(1.5)} = \frac{20}{3} m, in agreement. Note how much this is below the flat-ground Rmax=u2g=10R_{max} = \frac{u^{2}}{g} = 10 m. Brain-Teaser 2: For what launch angle α\alpha (measured from an incline of angle β\beta) does a projectile hit the incline at right angles? Evaluate for β=45∘\beta = 45^{\circ}, and explain why α=90∘\alpha = 90^{\circ} is not the answer even though a perpendicular launch sounds symmetric. Solution: Perpendicular impact requires the along-slope velocity to be zero at landing: vx′=ucos⁡α−gsin⁡β T=0v_{x'} = u\cos\alpha - g\sin\beta\,T = 0 with T=2usin⁡αgcos⁡βT = \frac{2u\sin\alpha}{g\cos\beta}. Substituting, ucos⁡α=2usin⁡αsin⁡βcos⁡βu\cos\alpha = \frac{2u\sin\alpha\sin\beta}{\cos\beta}, so cot⁡α=2tan⁡β\cot\alpha = 2\tan\beta. For β=45∘\beta = 45^{\circ}, cot⁡α=2\cot\alpha = 2, so tan⁡α=12\tan\alpha = \frac{1}{2} and α=arctan⁡(0.5)≈26.57∘\alpha = \arctan(0.5) \approx 26.57^{\circ}. Why is the answer not the symmetric-looking α=90∘\alpha = 90^{\circ} (launch perpendicular to the incline)? That launch does start with zero along-slope velocity, but gsin⁡βg\sin\beta immediately accelerates the body down the slope, so at landing its along-slope velocity is 0−gsin⁡β T=−2utan⁡β0 - g\sin\beta\,T = -2u\tan\beta, which is not zero; it lands some way down the slope from the launch point and strikes it obliquely. The genuinely symmetric-sounding conditions are three different things: perpendicular impact needs cot⁡α=2tan⁡β\cot\alpha = 2\tan\beta, returning to the launch point needs cot⁡α=tan⁡β\cot\alpha = \tan\beta (a vertical throw), and perpendicular launch is simply α=90∘\alpha = 90^{\circ}. The lesson is that along the slope the motion is uniformly accelerated, not uniform. Brain-Teaser 3: Two identical projectiles are fired with the same speed uu from the foot of a 30∘30^{\circ} incline, one up the slope and one down it, each at its own optimal angle. Find the ratio of the two maximum ranges and explain the physical reason for the asymmetry. Solution: Up-slope: Rup=u2g(1+sin⁡β)=u2g(1.5)=2u23gR_{up} = \frac{u^{2}}{g(1+\sin\beta)} = \frac{u^{2}}{g(1.5)} = \frac{2u^{2}}{3g}. Down-slope: Rdown=u2g(1−sin⁡β)=u2g(0.5)=2u2gR_{down} = \frac{u^{2}}{g(1-\sin\beta)} = \frac{u^{2}}{g(0.5)} = \frac{2u^{2}}{g}. The ratio is RdownRup=1+sin⁡β1−sin⁡β=1.50.5=3\frac{R_{down}}{R_{up}} = \frac{1+\sin\beta}{1-\sin\beta} = \frac{1.5}{0.5} = 3. So the downhill throw goes three times as far. Physically, in the rotated frame the along-slope acceleration gsin⁡βg\sin\beta opposes the motion when throwing uphill and assists it when throwing downhill, and it acts for the same flight time in both cases; the change of sign of that one term is the entire source of the asymmetry. A useful sanity check: setting β=0\beta = 0 makes the ratio 11, and letting β→90∘\beta \to 90^{\circ} makes Rdown→∞R_{down} \to \infty, which is the (correct) statement that throwing off a vertical cliff has no range limit.

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DPP: Projectile Motion on an Inclined Plane

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