Projectile Motion Calculator
Calculate the time of flight, maximum height, horizontal range and impact speed of a projectile from its initial velocity, launch angle, initial height and gravity. Supports m/s, km/h, ft/s, mph and different gravities.
Find Range, Maximum Height and Flight Time
Projectile Motion Formula
Where:
- \(v_0\) — initial velocity (speed at launch)
- \(\theta\) — launch angle above the horizontal
- \(h_0\) — initial height above the ground
- \(g\) — gravitational acceleration (9.81 m/s² on Earth)
- \(t\) — total time of flight
- \(h_{max}\) — maximum height reached
- \(R\) — horizontal range
- \(v_f\) — impact speed at the moment of landing
These are the standard ideal-projectile equations: the time of flight comes from solving the vertical motion equation \(h(t) = h_0 + v_0 \sin\theta \cdot t - \tfrac{1}{2} g t^2 = 0\), and the other quantities follow from the constant horizontal velocity and the vertical acceleration.
Projectile Motion – Calculation Example
Example: a ball is kicked from the ground at 20 m/s with a launch angle of 35°, under standard Earth gravity (\(g = 9.81\ \text{m/s}^2\), \(h_0 = 0\)).
\[ v_{0x} = 20 \cos 35^\circ = 16.383\ \text{m/s}, \qquad v_{0y} = 20 \sin 35^\circ = 11.472\ \text{m/s} \]
\[ t = \frac{2 \cdot 11.472}{9.81} = 2.339\ \text{s} \]
\[ h_{max} = \frac{11.472^2}{2 \cdot 9.81} = 6.71\ \text{m}, \qquad R = 16.383 \cdot 2.339 = 38.32\ \text{m} \]
Result: the ball stays in the air for about 2.339 s, reaches a maximum height of 6.71 m, and lands 38.32 m away, hitting the ground at 20 m/s (the same speed it was launched with, at 35° below the horizontal).
How to use the calculator
- Enter the Initial velocity (v₀) and pick its unit — m/s, km/h, ft/s or mph.
- Enter the Launch angle (θ) in degrees or radians. It must be between 0° and 90° above the horizontal.
- Set the Initial height (h₀) — leave it at 0 if the object is launched from the ground. Choose m or ft.
- Adjust Gravity (g) if needed. The default 9.81 m/s² applies to Earth; use 1.62 m/s² for the Moon or 3.72 m/s² for Mars.
- Read the results: time of flight, maximum height, horizontal range, impact speed and impact angle, all in your chosen units.
What is projectile motion?
Projectile motion is the motion of an object thrown into the air that is subject only to gravity (neglecting air resistance). The key insight, first worked out by Galileo, is that the horizontal and vertical motions are independent of each other: horizontally the object moves at a constant speed, while vertically it decelerates on the way up, momentarily stops at the peak, and accelerates downward on the way down.
This independence is what makes projectile problems so clean to solve. The horizontal distance covered depends only on the horizontal velocity component \(v_0 \cos\theta\) and the time in the air, while the height and flight time depend only on the vertical component \(v_0 \sin\theta\), the initial height and gravity. Combining the two gives the familiar parabolic trajectory.
The launch angle has a strong effect on the results. With the same speed, a 45° angle maximizes the range on level ground, while lower angles give a flatter, faster trajectory and higher angles give more height and hang time. An initial height above the landing level always increases both the range and the flight time. For a deeper dive into the underlying equations, see our kinematic equations solver.
Assumptions and limitations
- No air resistance. The calculator assumes an ideal vacuum. For real objects — especially fast, light or blunt ones like balls, arrows or bullets — air drag shortens the range noticeably, and the optimum launch angle drops below 45°.
- Flat ground, uniform gravity. The landing surface is assumed to be at height zero and gravity is treated as constant over the whole flight, which is an excellent approximation for everyday heights.
- No wind or spin. Headwinds, tailwinds and effects like the Magnus force from spin (which curves a football or baseball) are not included.
- Launch angle 0–90°. Only upward launches are accepted; downward angles (below the horizontal) are not supported.
Practical tips
To get the maximum range on level ground (\(h_0 = 0\)), set the launch angle to 45°. For throws from a height, the angle that maximizes distance is lower than 45° — try a few values and compare the ranges. Note that when \(h_0 = 0\), the impact speed always equals the launch speed and the impact angle equals the launch angle, a nice symmetry of ideal projectile motion.
If you also need velocities at intermediate points of the trajectory, the final velocity calculator and the displacement calculator can handle the individual vertical and horizontal segments, and the kinetic energy calculator shows how the projectile's energy depends on its speed.
Frequently Asked Questions
How do I calculate projectile range by hand without a calculator?
Split the launch velocity into components: vertical speed is \(v_0 \sin\theta\) and horizontal speed is \(v_0 \cos\theta\). For a launch from the ground, the flight time is \(t = 2 v_0 \sin\theta / g\), and the range is simply \(R = v_0^2 \sin(2\theta) / g\). For example, at 20 m/s and 35°, \(R = 400 \cdot \sin 70° / 9.81 \approx 38.3\) m. If the launch height is not zero, you must solve the quadratic \(h_0 + v_{0y} t - \tfrac{1}{2} g t^2 = 0\) for the time first.
Why does a 45° angle give the maximum range, and when is it not 45°?
On level ground, the range formula \(R = v_0^2 \sin(2\theta)/g\) is largest when \(\sin(2\theta) = 1\), which happens at \(\theta = 45°\). This only holds when the launch and landing heights are equal. If you throw from a cliff or elevated position, the optimal angle is lower than 45° — for example, around 35–40° for a throw from a 2 m height. With significant air resistance, the optimal angle also drops, typically to 30–40° for balls and other blunt objects.
Does the projectile hit the ground at the same speed it was launched?
Yes — but only when the launch and landing heights are the same. In ideal projectile motion without air resistance, energy is conserved: the projectile trades speed for height going up and gains it all back coming down. So with \(h_0 = 0\), the impact speed equals the launch speed, and the impact angle mirrors the launch angle (e.g., launched at 35° above horizontal, lands at 35° below). If the projectile lands lower than it started, it arrives faster; if it lands higher, slower.
Why does my hand calculation not match the calculator result?
The most common cause is a degrees vs. radians mix-up: your calculator or spreadsheet may be in radian mode while you typed degrees. Second, check unit consistency — if velocity is in km/h, convert it first (divide by 3.6 to get m/s; for mph multiply by 0.44704). Third, make sure you used the correct time of flight: with a nonzero initial height, the simple formula \(t = 2 v_0 \sin\theta / g\) no longer applies, and you must use the quadratic solution.
How do I handle a horizontal launch (thrown straight off a cliff)?
Set the launch angle to 0°. Then the vertical velocity starts at zero, and the object falls while moving forward at constant speed. From height \(h_0\), the fall time is \(t = \sqrt{2 h_0 / g}\) — about 1.43 s for a 10 m drop on Earth — and the range is simply \(v_0 \cdot t\). This is a classic physics problem, and the calculator handles it exactly the same way as any other angle.
How accurate are the results for real objects like balls, arrows or bullets?
The calculator models an ideal projectile with no air resistance, so it overestimates the range and flight time for real objects — the faster and less aerodynamic the object, the bigger the error. A football kicked at 20 m/s might lose 20–30% of its theoretical range to drag; a bullet can fall short by far more. Treat the results as an upper bound and expect real distances to be noticeably shorter, with a lower optimal launch angle than 45°.
Can I use this calculator for the Moon, Mars or other planets?
Yes — just change the gravity field. The Moon's gravity is about 1.62 m/s² and Mars's is 3.72 m/s², both of which dramatically change the results: the same 20 m/s launch at 35° flies roughly 232 m on the Moon versus 38 m on Earth, because range is inversely proportional to \(g\). The calculator keeps your chosen length and speed units, so you can run the same scenario under different gravity to compare.