Projectile Motion Calculator
Enter the initial velocity, launch angle, and (optionally) launch height. The calculator computes the velocity components, time to maximum height, maximum height, total time of flight, and horizontal range — showing the formula and the values plugged into it at every step. For the why behind these formulas, the Projectile Motion lesson walks through the derivation.
A projectile is anything moving under gravity alone. Break the initial velocity \(v_0\) into a horizontal component \(v_{0x} = v_0 \cos\theta\) and a vertical component \(v_{0y} = v_0 \sin\theta\), then solve the horizontal motion (constant velocity) and the vertical motion (constant acceleration of \(-g\)) separately. Time is the shared variable that links the two.
What This Calculator Computes
Given the initial velocity, launch angle, and launch height, this tool returns five quantities:
| Quantity | Formula |
|---|---|
| Horizontal component \(v_{0x}\) | \(v_0 \cos\theta\) |
| Vertical component \(v_{0y}\) | \(v_0 \sin\theta\) |
| Time to maximum height | \(t_{\text{up}} = \dfrac{v_{0y}}{g}\) |
| Maximum height | \(y_{\text{max}} = y_0 + \dfrac{v_{0y}^2}{2g}\) |
| Total time of flight | quadratic in \(t\); see below |
| Horizontal range | \(R = v_{0x} \cdot t_{\text{flight}}\) |
For a launch from ground level (\(y_0 = 0\)) the total time of flight is exactly \(2 t_{\text{up}}\) because the arc is symmetric. For a launch from above (\(y_0 > 0\)) or below the landing point, the calculator solves the quadratic \(y(t) = 0\) directly.
Worked Examples
Example 1: thrown ball
\(v_0 = 30\) m/s, \(\theta = 40°\), \(y_0 = 0\):
- \(v_{0x} = 30 \cos 40° \approx 22.98\) m/s
- \(v_{0y} = 30 \sin 40° \approx 19.28\) m/s
- \(t_{\text{up}} = 19.28 / 9.8 \approx 1.97\) s
- \(y_{\text{max}} = 0 + 19.28^2 / 19.6 \approx 18.95\) m
- \(t_{\text{flight}} = 2 t_{\text{up}} \approx 3.94\) s
- \(R = 22.98 \times 3.94 \approx 90.5\) m
Example 2: launched from a cliff
\(v_0 = 12\) m/s, \(\theta = 0°\), \(y_0 = 45\) m (launched horizontally off a cliff):
\(v_{0y} = 0\), so the time of flight comes from \(45 = \tfrac{1}{2}(9.8)t^2\), giving \(t \approx 3.03\) s. The range is \(12 \times 3.03 \approx 36.4\) m.
Tips for Using the Calculator
- Launch angle is measured from the horizontal: \(0°\) is a flat throw, \(90°\) is straight up.
- For a horizontal launch from a height (like a ball rolling off a table), use \(\theta = 0\) and set \(y_0\) to the launch height.
- Range is measured from the launch point to where the projectile crosses \(y = 0\). If \(y_0 > 0\), the projectile lands farther away than it would from ground level.
- Air resistance is ignored. For dense projectiles at moderate speeds this is a good approximation; for feathers or fast baseballs, less so.
Frequently Asked Questions
Why does maximum range happen at 45°?
For a launch from ground level, the range formula simplifies to \(R = v_0^2 \sin(2\theta) / g\). The factor \(\sin(2\theta)\) is maximized when \(2\theta = 90°\), i.e., when \(\theta = 45°\). For launches from a height, the optimal angle is slightly less than 45° because the extra time aloft favors more horizontal travel.
What if I want the projectile to land at a different height than it launched from?
Set \(y_0\) to the difference between launch and landing heights. If the projectile launches from a 10 m cliff and lands at the base, use \(y_0 = 10\). If it launches from ground and lands on a 5 m platform, use \(y_0 = -5\) (so the equation \(y(t) = 0\) corresponds to the landing height being zero).
Why does the angle have to be between 0° and 90°?
Angles outside this range either point the projectile downward at launch (between \(-90°\) and \(0°\)) or backward (above \(90°\)). Both are physically possible but require a different formula setup. This calculator assumes the standard case: launched forward and at or above horizontal.
Does this account for air resistance?
No. All formulas here assume the only force on the projectile is gravity. Real projectile trajectories with significant air resistance are not perfect parabolas and don't admit a simple closed-form solution — they're usually simulated numerically.
What if I'm on the Moon or some other planet?
Change the gravity field at the top of the form. The Moon is about \(1.625\) m/s²; Mars is about \(3.71\) m/s²; Jupiter is about \(24.8\) m/s². The rest of the math is the same.