Visual Projectile Motion & Ballistic Trajectory Kinematics Lab
Ideal projectile motion superposes constant horizontal velocity $x(t) = v_0\cos\theta \cdot t$ with uniform gravitational acceleration $y(t) = h_0 + v_0\sin\theta \cdot t - \frac{1}{2}gt^2$, tracing the parabolic trajectory $y(x) = h_0 + x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta}$.
y(x) = h₀ + x·tan θ − [g·x²] / [2·v₀²·cos²θ]Inputs & Parameters (4)
20-Row Comparison Table
| # | Launch Speed (v₀) (m/s) | Horizontal Range (R) | Peak Height | Flight Time |
|---|---|---|---|---|
| 01 | 5 m/s | 2.551 m | 0.64 m | 0.72 s |
| 02 | 10 m/s | 10.204 m | 2.55 m | 1.44 s |
| 03 | 15 m/s | 22.959 m | 5.74 m | 2.16 s |
| 04 | 20 m/s | 40.816 m | 10.2 m | 2.89 s |
| 05 | 25 m/s | 63.776 m | 15.94 m | 3.61 s |
| 06 | 30 m/s | 91.837 m | 22.96 m | 4.33 s |
| 07 | 35 m/s | 125 m | 31.25 m | 5.05 s |
| 08 | 40 m/s (Current) | 163.265 m | 40.82 m | 5.77 s |
| 09 | 45 m/s | 206.633 m | 51.66 m | 6.49 s |
| 10 | 50 m/s | 255.102 m | 63.78 m | 7.22 s |
| 11 | 55 m/s | 308.673 m | 77.17 m | 7.94 s |
| 12 | 60 m/s | 367.347 m | 91.84 m | 8.66 s |
| 13 | 65 m/s | 431.122 m | 107.78 m | 9.38 s |
| 14 | 70 m/s | 500 m | 125 m | 10.1 s |
| 15 | 75 m/s | 573.98 m | 143.49 m | 10.82 s |
| 16 | 80 m/s | 653.061 m | 163.27 m | 11.54 s |
| 17 | 85 m/s | 737.245 m | 184.31 m | 12.27 s |
| 18 | 90 m/s | 826.531 m | 206.63 m | 12.99 s |
| 19 | 95 m/s | 920.918 m | 230.23 m | 13.71 s |
| 20 | 100 m/s | 1,020.408 m | 255.1 m | 14.43 s |
Key Analytical Takeaways
- 01On level ground ($h_0 = 0$), maximum horizontal range $R_{\max} = v_0^2 / g$ occurs at $\theta = 45^\circ$, and complementary angles ($30^\circ$ and $60^\circ$) land at the exact same spot.
- 02At the trajectory apex, vertical velocity $v_y = 0$, leaving purely horizontal speed $v_x = v_0\cos\theta$.
- 03Slide Gravity $g$ between Moon ($1.6\text{ m/s}^2$), Mars ($3.7\text{ m/s}^2$), Earth ($9.8\text{ m/s}^2$), and Jupiter ($24.8\text{ m/s}^2$) to compare planetary arcs.
Frequently Asked Questions
Why does maximum range happen below 45° when launching from an elevated cliff (h₀ > 0)?+
Because the extra drop h₀ gives the projectile additional air time regardless of upward launch velocity, trading a shallower angle for higher horizontal speed v₀·cos(θ) extends total range.
Why do 30° and 60° launch angles produce the same range on flat ground?+
Level-ground range is proportional to sin(2θ). Since sin(2×30°) = sin(60°) and sin(2×60°) = sin(120°) are equal, any complementary pair θ and (90° − θ) shares the same range.
Does 45° still maximize horizontal range when launching from an elevated cliff (h₀ > 0)?+
No! When launching from initial height h₀ > 0 above the landing plane, the optimal launch angle drops below 45°—specifically θ_opt = arcsin(v₀ / √(2v₀² + 2gh₀))—trading apex height for extra forward velocity during the extended fall.