Maximum Height Projectile Motion Calculator

Find the maximum height, time-to-peak, total flight time, and range of a projectile from initial speed and launch angle.

Science Earth · Moon · Mars Range + time
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H = v₀²sin²(θ) / (2g)

Projectile motion · multi-planet g · range and time

Instructions — Maximum Height Projectile Motion Calculator

1

Enter speed and angle

Initial velocity v₀ in m/s. Launch angle θ in degrees, 0 (horizontal) to 90 (straight up). Default of 20 m/s at 45° gives 10.2 m peak.

2

Set initial height (optional)

If the projectile starts above ground (a tower, platform, hand height), enter h₀. Max height becomes h₀ + relative peak.

3

Switch planets

Use the gravity dropdown for Moon, Mars, Jupiter. Same v₀ and θ on the Moon gives roughly 6× the height of Earth.

Formulas

Maximum Height
$$ H = \frac{v_0^2 \sin^2\theta}{2g} $$
Peak above the launch point. Depends only on the vertical velocity component v₀sin(θ).
Time to Peak
$$ t_{peak} = \frac{v_0 \sin\theta}{g} $$
Vertical velocity becomes zero at the peak. Half of the total flight time when launched from ground level.
Horizontal Range
$$ R = \frac{v_0^2 \sin(2\theta)}{g} $$
Distance traveled horizontally (when h₀ = 0). Max range at θ = 45°. Sin(2θ) means 30° and 60° give the same range.
Total Flight Time
$$ t_f = \frac{v_0 \sin\theta + \sqrt{(v_0\sin\theta)^2 + 2gh_0}}{g} $$
Time from launch to landing, accounting for launch height h₀. Reduces to 2v₀sin(θ)/g when h₀ = 0.
Trajectory
$$ y = h_0 + x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta} $$
The path is a parabola. The negative x² term gives the downward curve.
Speed at Peak
$$ v_{peak} = v_0 \cos\theta $$
At maximum height the vertical velocity is zero. Only horizontal component remains.

Reference

Maximum Height for v₀ = 20 m/s on Earth
Angle θH_max (m)Range (m)Flight time (s)
15°0.8620.41.06
30°5.1035.32.04
45°10.2040.82.88
60°15.3035.33.53
75°19.0420.43.94
90°20.3904.08

Gravitational acceleration by planet

All values in m/s².

BodygRelative to Earth
Earth9.811.00 ×
Moon1.620.17 ×
Mars3.720.38 ×
Venus8.870.90 ×
Jupiter24.792.53 ×
Mercury3.700.38 ×

Article — Maximum Height Projectile Motion Calculator

Maximum Height Projectile Motion Calculator

Maximum height in projectile motion is H = v₀² × sin²(θ) / (2g). A ball thrown at 20 m/s at 45° on Earth reaches 10.2 m — about a three-story building. The same throw at 60° goes 50% higher because sin² grows faster than the loss in horizontal velocity matters for vertical reach.

What maximum height means in projectile motion

A projectile is anything in motion under gravity alone after release — a thrown ball, a fired arrow, a kicked soccer ball. Air resistance is normally neglected in the ideal model. The projectile's vertical velocity starts as v₀ sin(θ) upward, slows under gravity, hits zero at the peak, and then becomes downward as it falls. The peak is the maximum height.

Horizontal velocity is constant the whole time (still no air resistance, no friction). So at the peak the projectile is still moving forward at v₀ cos(θ) but with zero vertical velocity. That is the instant of maximum height. Galileo first separated the motion into independent horizontal and vertical components around 1600 — the insight that makes projectile motion tractable.

Did you know

Galileo's Two New Sciences (1638) proved that the projectile path is a parabola. Aristotle had taught for nearly two thousand years that projectiles travel in a triangular path — straight forward, then straight down — and Galileo's parabolas overturned that received wisdom in a single page of geometry.

The maximum height projectile formula

The peak equation is H = v₀² sin²(θ) / (2g). Three inputs: initial speed, launch angle, gravity. One output: height above launch point. If the projectile starts at h₀ above ground, the absolute maximum height is h₀ + H.

Two related quantities follow immediately. Time to peak: t_peak = v₀ sin(θ) / g. Total flight time (for h₀ = 0): t_flight = 2v₀ sin(θ) / g — exactly twice the peak time, because the trajectory is symmetric. Horizontal range: R = v₀² sin(2θ) / g, which peaks at θ = 45°.

Earth, v₀ = 20 m/s, h₀ = 0
15° H = 1.37 m
30° H = 5.10 m
45° H = 10.20 m
60° H = 15.30 m
75° H = 19.04 m
90° H = 20.39 m

How launch angle changes max height

Maximum height depends entirely on the vertical component of launch velocity. Horizontal velocity has no influence — it does not interact with gravity. So sin²(θ) drives the height curve: 0 at θ = 0, peaks at θ = 90° (straight up). At 45° you get sin² = 0.5, so half of the maximum-possible peak; at 60° you get sin² = 0.75, three-quarters.

Range tells a different story. R uses sin(2θ), which peaks at θ = 45°. The trade-off is built in: low angles give long range but low peak; high angles give high peak but short range. A 45° launch is the sweet spot for maximum range, not maximum height.

Max height
θ = 90°
straight up
Max range
θ = 45°
flat ground

Maximum height for different speeds

Height scales with v₀². Double the speed, the max height quadruples. A baseball pitcher throws fastballs around 40 m/s; vertically (and ignoring drag) that would peak at 81 m. A soccer goal kick averages 25 m/s and peaks around 32 m. A golf drive leaves the clubface at roughly 75 m/s; without air resistance the ball would rise above 280 m, but drag and lift cut it to a real peak near 30–50 m.

Vertical launches are the simplest case: θ = 90°, sin = 1, so H = v₀² / 2g. A stone thrown straight up at 10 m/s reaches 5.1 m. The 30 m/s rocket fired straight up clears 45.9 m before falling back. Equation is symmetric: time up equals time down, and impact speed equals launch speed.

Projectile motion on other planets

Lower gravity, higher peaks. The relationship is inversely proportional: H scales as 1/g for fixed v₀ and θ. Moon gravity (1.62 m/s²) is 16.5% of Earth's, so a throw on the Moon goes 6.1× higher. Mars gravity (3.72) is 38% of Earth's, so a Mars throw goes 2.6× higher. Jupiter at 24.8 is 2.5× Earth's, so a throw there reaches only 40% of its Earth peak.

The famous Apollo 14 golf shot illustrates this. Alan Shepard's one-handed swing with a 6-iron, hampered by his pressurized suit, sent the ball "miles and miles" in his words — in reality, modern analysis of the footage suggests about 40 yards, still far more than the same swing would produce on Earth where air drag plus the suit constraint would have killed it.

Tip

To estimate maximum height fast, use H ≈ v₀y² / 20 in SI units (rounding g to 10). A 10 m/s vertical velocity gives H ≈ 5 m. The actual answer using g = 9.81 is 5.10 m — close enough for mental math.

Real-world projectile trajectories

Drag changes everything. A baseball thrown at 40 m/s at 45° "should" carry 163 m without air resistance. The actual carry distance is closer to 110 m because the drag force grows roughly with v² and pulls steeply on small high-speed objects. Lift from spin (Magnus effect) modifies the trajectory further; a backspun golf ball travels significantly farther than a no-spin shot.

For dense, slow-moving projectiles the no-drag model is accurate to within a few percent. Shot put (7.26 kg, low launch speed) is well-approximated by ideal projectile motion. Discus and javelin involve significant aerodynamic lift and are not. Bullets are nearly straight-line over short ranges but follow a noticeable arc at battlefield distances; military fire-tables embed precise drag corrections.

Common projectile motion mistakes

Three traps catch students and engineers alike.

Angle in radians vs. degrees

If you forget to convert degrees to radians before applying sin(), the answer is gibberish. JavaScript's Math.sin takes radians; spreadsheets like Excel take radians; physics textbooks usually quote degrees. Always confirm.

Second, conflating maximum height and total flight time. Time to peak is half of total flight time only when h₀ = 0. From a 10 m tower the projectile spends less time going up than coming down. Third, ignoring h₀ in the range formula. The simple sin(2θ)/g formula only works for ground-to-ground launches; otherwise use the full time-of-flight expression with the quadratic discriminant.

FAQ

H = v₀² × sin²(θ) / (2g). Maximum height depends only on the vertical component of initial velocity. Horizontal motion does not affect height. Example: 20 m/s at 45°: H = 400 × 0.5 / 19.62 = 10.2 m on Earth.
Range R = v₀² × sin(2θ) / g. The sin(2θ) factor maxes when 2θ = 90°, so θ = 45°. This balances horizontal velocity and time-of-flight perfectly. Launching from above ground level (h₀ > 0) shifts the optimum slightly above 45°.
t_peak = v₀ × sin(θ) / g. The vertical velocity reaches zero at the peak. For 20 m/s at 45° on Earth: t_peak = 20 × 0.707 / 9.81 = 1.44 s. Total flight time (for h₀ = 0) is exactly twice this.
No, in idealized projectile motion. The acceleration is g regardless of mass. A bowling ball and a ping-pong ball launched at the same v₀ and θ reach the same maximum height — in vacuum. Air resistance breaks the symmetry, slowing the lighter object faster.
Lunar gravity is 1.62 m/s² versus Earth's 9.81 — about 6.05× weaker. Maximum height and range both scale inversely with g. A throw that reaches 10 m on Earth reaches roughly 60.5 m on the Moon. Time-of-flight scales as 1/g too, so the lunar throw stays in the air about 6× longer.
This calculator ignores it. For dense projectiles (cannonball, bullet) at moderate speeds, air resistance changes the answer by 5–20%. For light objects (foam ball, paper plane) the effect is dominant. Real ballistics calculations include drag coefficient, air density, and projectile cross-section.
Because the launch height (h₀ ≈ 2 m) shifts the optimal angle below 45°. The shot is released above ground level, so the time-of-flight side of the range equation is already biased in favor of horizontal velocity. Elite throwers use 34–38°, validated by track-and-field biomechanics studies.
v_peak = v₀ × cos(θ). Vertical velocity is zero at the peak; only the horizontal component survives. For a 20 m/s, 45° launch: v_peak = 20 × 0.707 = 14.1 m/s — still moving forward, just not upward anymore.
Higher launch increases range. A baseball thrown from a 2 m mound travels farther than the same throw at ground level. The full range formula handles this: R = v₀cos(θ) × t_flight, where t_flight grows when h₀ > 0. The optimal angle also shifts above 45° as h₀ grows.