Projectile Motion: Launch Velocity from Range and Angle
Calculate the launch speed needed to hit a target at a known range and angle. Get required velocity, flight time, max height, and trajectory chart.
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Definition
The inverse projectile problem fixes the launch angle, target range, and (optionally) the launch height above the landing surface, then solves for the initial speed required to reach the target. It is the counterpart of the standard "given speed and angle, find range" problem and applies whenever launch geometry is constrained — body mechanics, weapon elevation limits, terrain — and the unknown is muzzle velocity, throw speed, or release force.
How It Works
Same-Height Case
If launch and landing are at the same height, the textbook range formula is:
Solving for :
The required speed diverges as or — a horizontal or vertical launch would need infinite speed to land at any positive distance — and it is minimized exactly at 45°.
With Initial Height
If the projectile launches from a height above the landing surface — a basketball release, a cliff-top throw, a cannon on a hill — the range equation gains an extra term and reduces to a quadratic in :
The required velocity is lower than the same-height case because gravity gets extra time to work on the projectile. The bigger relative to , the larger the saving.
Required Velocity vs. Launch Angle
For a fixed range and same-height launch:
| Angle | Required v₀ (R = 100 m, g = 9.81) | Notes |
|---|---|---|
| 15° | 44.3 m/s | Flat shot — high velocity |
| 30° | 33.7 m/s | |
| 45° | 31.3 m/s | Minimum velocity — the optimum |
| 60° | 33.7 m/s | Mirror of 30° |
| 75° | 44.3 m/s | High lob — same as flat shot |
Two angles equidistant from 45° require the same launch speed. The 45° angle gives the lowest possible velocity for a given range — the relevant case when minimizing launch effort or energy.
Worked Example
A basketball free throw has fixed geometry: 4.6 m to the basket, release height around 2.3 m, basket height 3.05 m, so the release is about 0.75 m below the rim. Players typically release at 50–55°. With m, , and m (negative because the basket is above the release point), the calculator returns roughly 7.3 m/s15 ft to the basket, release height around 7.5 ft, basket height 10 ft, so the release is about 2.5 ft below the rim. Players typically release at 50–55°. With m, , and m (negative because the basket is above the release point), the calculator returns roughly 7.3 m/s ≈ 24 ft/s — close to measured biomechanics data on NBA shooters.
Applications
The inverse range formula applies wherever launch geometry is fixed and the unknown is the speed needed to reach a target.
- Siege-engine sizing. A catapult or trebuchet has a release angle fixed by its build and a target distance set by the wall. The required velocity indicates how much potential energy the counterweight or torsion mechanism must deliver.
- Game-engine ballistics. For an AI archer that must hit a target, fixing the launch angle to a visually plausible value (45° for a high arc, 20° for a flat fast shot) and the target distance yields the velocity to pass to the projectile, removing the need for iterative tuning.
- Reverse-engineering a throw. From a slow-motion video of a baseball pitch — release point, the angle the ball leaves the hand, and the plate distance — the formula returns the release velocity, which is useful for coaching reviews where direct radar readings are unavailable.
Limits
- No air resistance. This is most significant for slow projectiles (basketballs, long-flight throws), where drag has a meaningful effect. The calculator gives the vacuum baseline; real-world launches typically need 5–25% more velocity.
- No spin or lift. The Magnus effect (curving balls), arrow fletching stabilization, and aerodynamic lift on long projectiles are not modeled.
- Launch angle constraints. must lie in $(0°, 90°)$ — the formulas degenerate at $0°$ and $90°$.
- Initial height limits. A negative (target above launch) requires that the trajectory actually reach the target height. When the chosen angle lacks the vertical capacity to do so, no real solution exists; a higher angle or a shorter range resolves it.
Frequently Asked Questions (FAQ)
Why do 30° and 60° require the same launch speed for the same range?
Because the range formula depends on sin(2θ), which is symmetric about 45°. sin(60°) = sin(120°), so a 30° launch (2θ = 60°) and a 60° launch (2θ = 120°) cover the same range at the same speed. The two trajectories differ — the 30° shot is flat and fast in flight time, the 60° shot is high and lobbed — but they need the same initial speed.
Why does a 45° launch use the least velocity?
sin(2θ) is maximized at 2θ = 90°, i.e. θ = 45°. For a fixed range R, the required v₀ is proportional to 1/√sin(2θ), so the maximum sin(2θ) gives the minimum v₀. 45° is the energy-efficient angle for any same-height projectile problem.
How does an initial height (cliff, balcony) change the required velocity?
A positive initial height (launching above the target) lowers the required velocity, because gravity has more time to deliver the projectile. A negative initial height (target above launcher, like a basketball hoop) raises the required velocity. The calculator solves the full equation including h₀.
My calculation returns no real solution — what does that mean?
You are asking for a target above launch (negative h₀) at an angle that cannot deliver enough vertical lift. Try a steeper launch angle, a shorter range, or accept that the geometry is impossible at any reasonable speed.
Disclaimer
This is a vacuum model and ignores air resistance, lift, wind, and Magnus effects. Real launches typically need 5–25% more velocity than the vacuum prediction; for engineering or sports analysis, use a drag-corrected model.