Projectile Motion: Velocity & Angle from Max Height and Range

Last updated: 2026-05-16

Inverse projectile motion

Inverse projectile motion is the problem of recovering the launch conditions — initial velocity and launch angle — from the trajectory they produce, specified here by a maximum height and a horizontal range. The forward problem starts from a known speed and angle and asks where the projectile lands; the inverse problem starts from a desired peak height and range and asks what initial velocity and launch angle are required to hit them. For any feasible pair of target values there is exactly one launch velocity and one launch angle that produce both simultaneously.

How the two targets fix the launch

The vacuum projectile model describes the motion with two relationships. For the symmetric ground-to-ground case (initial height $h_0 = 0$):

$H = \frac{v_0^2 \sin^2\theta}{2g} \qquad R = \frac{v_0^2 \sin(2\theta)}{g}$

where $H$ is maximum height, $R$ is range, $v_0$ is initial speed, $\theta$ is launch angle from horizontal, and $g$ is gravity. These are two equations in the two unknowns $v_0$ and $\theta$, so a known $(H, R)$ pair determines both.

Taking the ratio $H / R$ eliminates $v_0$:

$\frac{H}{R} = \frac{\sin^2\theta}{2 \sin(2\theta)} = \frac{\tan\theta}{4}$

which isolates the angle:

$\theta = \arctan\!\left(\frac{4H}{R}\right)$

Substituting $\theta$ back into the height equation gives the speed:

$v_0 = \sqrt{\frac{2gH}{\sin^2\theta}}$

The derivation reduces to two steps: solve for the angle from the height-to-range ratio, then solve for the speed.

How angle responds to the height-to-range ratio

The relation $\tan\theta = 4H/R$ has a direct geometric reading: the height of the trajectory's peak relative to the range sets the launch angle. The table below lists the angle for several ratios.

The 0.25 row recovers the familiar 45° result from the other direction: at 45° the apex sits exactly $R/4$ above the ground.

Worked example

Consider an arrow that must clear a 6 m wall and land 30 m away. With $H = 6$ and $R = 30$, the calculator returns $\theta \approx 38.7°$ and a velocity around 17.4 m/sa 20 ft wall and land 100 ft away. With $H = 20$ ft and $R = 100$ ft (or 6 m and 30 m), the calculator returns $\theta \approx 38.7°$ and a velocity around 17.5 m/s ≈ 57.4 ft/s (on Earth). The angle follows from $\arctan(4 \times 6 / 30) = \arctan(0.8)$, and the speed from substituting that angle into the height equation.

Variations and edge cases

• Non-zero initial height. When the launch sits above the landing surface — a cliff, a table edge, or a basketball release point — the maximum height is still measured from the landing surface, so it must be at least as large as the initial height. With $h_0 > 0$ the angle generalises to $\tan\theta = 2((H - h_0) + \sqrt{H(H - h_0)}) / R$; the simple $\tan\theta = 4H/R$ relation is the special case $h_0 = 0$. A higher launch reaches the same $H$ and $R$ with a flatter angle. For landings on a sloped surface rather than flat ground at a different elevation, use the incline calculator instead.
• Single solution per target. Each feasible $(H, R)$ pair maps to exactly one $\theta$ and one $v_0$. When $H/R$ is large — for instance $H > R/4$, which pushes $\theta$ past 45° — the math still resolves, but the trajectory becomes increasingly steep and, in reality, increasingly dominated by drag.

Applications and limits

The same target can be reached two ways — a flat fast trajectory or a high slow one — which makes the inverse view useful for teaching the range-versus-height trade-off: holding $R$ fixed and varying $H$ shows the angle $\theta$ climb from a flat shot toward a near-vertical lob, with velocity increasing in both directions away from the optimum 45°.

The inverse calculation also appears in reverse-engineering recorded motion. A long jumper's hang time and distance are public: hang time gives flight time and distance gives range, from which takeoff velocity and angle can be recovered and compared across athletes. Mike Powell's 1991 world-record long jump (8.95 m, ≈1.0 s in the air, peak ~0.5 m) yields a takeoff angle around 12.6° and ~14.4 m/s under the vacuum model. The model overstates speed and flattens the angle relative to tracked biomechanics data — a reminder of how much air resistance and the body's lift profile affect a real athlete.

Pre-1900 field-artillery doctrine relied on charts performing this same calculation by hand: given a desired ridgeline clearance (maximum height) and a target distance, the gunner read off elevation and powder charge. The vacuum approximation the calculator reproduces is the classical core of those tables, which then corrected heavily for drag, wind, and Earth rotation.

These results assume the vacuum model. Real projectiles deviate substantially: a baseball loses 20–40% of its vacuum range to drag, while a dense bullet loses much less by mass though still a measurable amount. For ballistics, sports analysis, or engineering work where precision matters, use a model that incorporates drag.