Choosing coordinates for a projectile with a sloped landing
A ball is launched from the origin with speed 20 m/s at angle θ above the horizontal, over horizontal ground, and lands on a plane inclined at 30° to the horizontal that rises away from the launch point. Take g = 10 m/s². A student sets up axes along and perpendicular to the inclined plane, resolves gravity into components -g sin30° along the plane and -g cos30° perpendicular to it, and writes the perpendicular displacement as zero at landing. A second student uses standard horizontal/vertical axes and sets y = x tan30° at landing. Both get different equations for the flight time and disagree about which is correct.
1. Show that both setups lead to the same condition on the flight time, and state that condition.
Read the worked answer
Standard axes. With x = 20cosθ · t and y = 20sinθ · t - 5t^2, landing on the incline means y = x tan30°:
20sinθ · t - 5t^2 = 20cosθ · t · tan30°.
For t > 0, divide by t: 20sinθ - 5t = 20cosθ tan30°, so
t = 4(sinθ - cosθ tan30°).
Tilted axes. Along the plane: s = 20cos(θ - 30°)·t - 5sin30°·t^2. Perpendicular: 0 = 20sin(θ - 30°)·t - 5cos30°·t^2, giving t = 4sin(θ - 30°)/cos30°.
Now 4sin(θ - 30°)/cos30° = 4[sinθ cos30° - cosθ sin30°]/cos30° = 4(sinθ - cosθ tan30°).
So both give t = 4(sinθ - cosθ tan30°). Neither is wrong; the tilted-axis version is just the perpendicular equation solved directly, and it is identical to the standard-axis condition after using a compound-angle identity.
2. For θ = 60°, compute the flight time and the distance travelled along the incline. Comment on whether the ball actually lands on the rising plane.
Read the worked answer
With θ = 60°: sinθ = √3/2, cosθ = 1/2, tan30° = 1/√3.
t = 4(√3/2 - (1/2)(1/√3)) = 4(√3/2 - 1/(2√3)) = 4 · (3 - 1)/(2√3) = 4 · 2/(2√3) = 4/√3 ≈ 2.31 s.
Distance along the incline: s = 20cos(60° - 30°)·t - 5sin30°·t^2 = 20cos30°·(4/√3) - 5(1/2)(16/3).
= 20(√3/2)(4/√3) - 40/3 = 40 - 40/3 = 80/3 ≈ 26.7 m.
Since t > 0 and s > 0, the ball does land on the rising plane. Note the check that θ = 60° exceeds the incline angle 30°, so the launch is directed above the plane - consistent with a positive perpendicular component of initial velocity.
3. Find the value of θ for which the ball lands furthest up the incline, and explain why the answer is not θ = 45°.
Read the worked answer
Using tilted axes, the perpendicular equation gives t = 4sin(θ - 30°)/cos30°, and s = 20cos(θ - 30°)t - 5sin30° t^2. Substituting and simplifying (standard result for range up an incline φ):
s = (u^2/g) · [sin(2θ - φ) - sinφ]/cos^2φ.
With u = 20, g = 10, φ = 30°: s = 40[sin(2θ - 30°) - 1/2]/(3/4) = (160/3)[sin(2θ - 30°) - 1/2].
This is maximised when sin(2θ - 30°) = 1, i.e. 2θ - 30° = 90°, so θ = 60°.
The answer is not 45° because the target is not horizontal. The launch must be biased upward relative to the horizontal to reach far up a rising plane; the optimum bisects the angle between the plane and the vertical, i.e. θ = (90° + φ)/2 = 60°. At θ = 45° the ball lands closer to the launch point.