Projectile from a Cliff
Projectile launched from elevated platform with adjustable height, angle, and speed; trajectory and impact point shown
Objective
Verify that a point-mass projectile launched from an elevated cliff follows x = v₀·cos θ·t and y = H + v₀·sin θ·t − ½g·t², with no air resistance or spin. Observe how cliff height independently extends flight time, how launch angle controls the arc shape, and how initial speed sets the horizontal range — each parameter varying while the others remain fixed.
Setup
- Set Height to 50 m, Angle to 30°, and Speed to 20 m/s — press Start and watch the amber arc form from the cliff edge to the ground.
- Record the Horiz. dist. (m) and Time (s) readouts the moment the red × appears at impact. The expected range is ≈ 75.7 m at ≈ 4.37 s.
- Press Reset, change Height to 100 m (keep Angle 30°, Speed 20 m/s), press Start again — note how the longer fall extends flight time and horizontal distance.
- Press Reset, set Angle to 0° (horizontal launch from H = 50 m, Speed 20 m/s), press Start — compare the flatter arc and shorter time to the 30° run.
Analytical Prediction
For H = 50 m, v₀ = 20 m/s, θ = 30°, g = 9.8 m/s², the vertical launch component is vy₀ = 20·sin 30° = 10 m/s and the horizontal component is vx = 20·cos 30° ≈ 17.32 m/s. Setting y(t) = 0 gives the quadratic 4.9t² − 10t − 50 = 0:
Horizontal range:
The apex occurs at t* = vy₀/g = 10/9.8 ≈ 1.02 s, reaching height H + vy₀²/(2g) = 50 + 100/19.6 ≈ 55.1 m above ground. The Horiz. dist. readout should read ≈ 75.7 m and Time ≈ 4.37 s at landing.
Results Analysis
Run the default configuration (H = 50 m, θ = 30°, v₀ = 20 m/s) and let the sim reach natural stop. Read the four HUD boxes at the moment the red × appears: Time (s) should be ≈ 4.37 s, Horiz. dist. (m) should be ≈ 75.7 m, Height (m) should read 0.0, and Speed (m/s) should equal the impact speed sqrt(vx² + vy_impact²) ≈ sqrt(17.32² + (10 − 9.8·4.37)²) ≈ 36.4 m/s. Compare the Horiz. dist. readout against the analytical 75.7 m — typical residual is under 0.5 m due to the fixed-substep integrator. For the H = 100 m run, the predicted range is ≈ 97.8 m at ≈ 5.65 s; verify both readouts fall within 1 m and 0.05 s of those values.
Source of Error
This simulation models the projectile as a point mass in vacuum under uniform gravity (g = 9.8 m/s²). It omits air drag, the Magnus effect (spin-induced lift), wind, the Coriolis effect, and any variation of g with altitude. The analytical prediction in the Prediction section uses the same point-mass, no-drag idealization, so both share identical physical assumptions — the idealizations cancel and do not contribute to the residual between predicted and measured values. The residual gap between the Horiz. dist. readout and the analytical range is therefore purely numerical, not physical, for this sim.
Further Exploration
- Set Angle to 0° (horizontal) with H = 50 m, v₀ = 20 m/s — does the horizontal launch give a longer or shorter range than θ = 30°? Why does a positive angle help even though it costs some horizontal velocity component?
- Try Angle = −20° (downward launch) at the same H and v₀. Does flight time shrink compared to θ = 0°? Estimate the ratio of flight times analytically using t = sqrt(2H/g) for the horizontal case and the full quadratic for the downward case.
- Compare H = 10 m and H = 100 m at fixed θ = 0°, v₀ = 20 m/s. Flight time scales as sqrt(H) — does doubling H from 25 m to 100 m double the time, or increase it by a factor of sqrt(4) = 2?
- At H = 50 m, v₀ = 20 m/s, sweep the Angle slider from −30° to 60° and watch the range readout. Find the angle that maximises range — is it 45° as in the level-ground case, or does the elevated launch shift the optimum toward a smaller angle?