Maximum Height Calculator - Projectile Motion
Find how high a projectile climbs, when it reaches the top and how fast it moves there, or work out the launch speed needed for a target height.
🏔️ What is the Maximum Height of a Projectile?
The maximum height of a projectile is the highest point of its flight, the apex, where it stops climbing and starts to fall. A projectile launched at speed v₀ and angle θ above the horizontal starts with a vertical velocity v₀ sin(θ). Gravity removes that vertical velocity at a steady rate g, so the climb ends when the vertical velocity reaches zero. Everything about the apex follows from that one fact: the height H = (v₀ sin θ)² / (2g) above the launch point, the time t = v₀ sin(θ) / g to reach it, and the horizontal distance already covered, v₀ cos(θ) × t.
People use the apex calculation in many places. Coaches estimate how high a kicked football or a punted ball will go (hang time is twice the time to the apex when landing at the launch level). Engineers size fountain jets, sprinkler throws and water cannons from the height they must reach. Safety planners check how high a falling object or a launched flare can climb. Students use it to practise vertical-motion kinematics and to see why a ball thrown on the Moon rises six times higher than the same throw on Earth. This page also runs the problem backwards: give it the height you need and the angle you will use, and it returns the launch speed.
This page is about the apex only, which is why it focuses on the vertical motion and the energy view. The kinetic energy of the vertical component, ½ (v₀ sin θ)² per kilogram, is converted completely into gravitational potential energy g × H at the top, which is the same result as the kinematics formula obtained by conservation of energy. The horizontal part of the velocity is not touched by gravity, so at the apex the projectile still moves at v₀ cos(θ). For the full flight, including range and landing, use the projectile motion calculator. For the underlying constant-acceleration equations, see the kinematic equations calculator.
Two points are easy to get wrong. First, the speed at the apex is not zero for an angled launch: only the vertical part vanishes, and the horizontal part remains. Only a vertical throw (90 degrees) is momentarily at rest at the top. Second, mass does not appear in the formula. A heavy ball and a light ball launched with the same velocity rise to the same height in the ideal model. Real objects also feel air resistance, which lowers the apex, so treat the results as the ideal vacuum value and an upper bound for fast or light objects.
📐 Formula
📖 How to Use This Calculator
Steps
💡 Example Calculations
Example 1 — Baseball Hit at 40 Degrees
A baseball leaves the bat at 40 m/s (about 89 mph) at 40° from 1 m above the ground, on Earth
Example 2 — Vertical Throw on Earth
A ball is thrown straight up (90°) at 20 m/s from the ground on Earth
Example 3 — The Same Vertical Throw on the Moon
The same 20 m/s vertical throw, but with lunar gravity g = 1.62 m/s²
Example 4 — Ball Kicked from a 10 m Roof
A ball is kicked at 18 m/s and 35° from a roof 10 m above the ground, on Earth
Example 5 — Fountain Jet Reaching 20 m
What exit speed does a vertical fountain jet need to reach 20 m on Earth (90°)?
❓ Frequently Asked Questions
🔗 Related Calculators
What is the formula for the maximum height of a projectile?
The rise above the launch point is H = (v0 sin theta)^2 / (2g), where v0 is the launch speed, theta is the launch angle above the horizontal and g is gravitational acceleration. Add the launch height h0 to get the height above the ground. A 40 m/s launch at 40 degrees on Earth rises 33.7 m.
How do you find the time to reach maximum height?
The time to the apex is t = v0 sin(theta) / g, because the vertical velocity falls from v0 sin(theta) to zero at a steady rate g. For 20 m/s straight up on Earth, t = 20 / 9.80665 = 2.04 s. The fall back to the launch level takes the same time.
What is the speed of a projectile at its maximum height?
At the apex the vertical velocity is zero, so the speed equals the horizontal component v0 cos(theta). It is not zero unless the launch angle is 90 degrees. A 40 m/s launch at 40 degrees has a speed of 30.64 m/s at the top.
What launch angle gives the greatest maximum height?
90 degrees, a vertical throw, gives the greatest height for a fixed launch speed, H = v0^2 / (2g). The height scales with sin squared of the angle, so 45 degrees gives half of the vertical-throw height and 30 degrees gives a quarter.
Does the mass of the projectile affect its maximum height?
No. In the ideal model without air resistance, mass cancels out of the height formula, so a heavy ball and a light ball launched with the same velocity peak at the same height. Mass matters only for the energy involved, not the height.
How high does a ball go on the Moon compared with Earth?
The height is inversely proportional to g. The Moon's surface gravity is about 1.62 m/s2 against 9.80665 m/s2 on Earth, so the same throw rises about 6.05 times higher on the Moon. A 20 m/s vertical throw reaches 20.4 m on Earth and 123.5 m on the Moon.
How do I calculate the launch speed needed to reach a certain height?
Rearrange the formula: v0 = sqrt(2 g H) / sin(theta). To rise 20 m on Earth with a vertical throw you need sqrt(2 x 9.80665 x 20) = 19.81 m/s. At smaller angles you need more speed because only part of it is vertical.
Why does the calculator reject a launch angle of 0 degrees in the speed mode?
At 0 degrees the launch has no vertical component, so no speed can lift the projectile above its launch height. The formula divides by sin(0), which is zero, so the required speed is infinite. Enter an angle above 0 degrees.
How does launch height change the maximum height?
The rise above the launch point does not depend on launch height. The height above the ground is simply h0 plus the rise. A 10 m starting height adds exactly 10 m to the apex above the ground.
Is maximum height the same as range?
No. Maximum height is the highest point of the flight. Range is the horizontal distance to landing. Height depends on sin squared of the angle, range on sin of twice the angle, so the 45 degree angle that maximizes range is not the angle that maximizes height.