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Maximum Height Calculator — Peak Height of a Projectile

Find the maximum height of a projectile using the formula h_max = h₀ + (v₀·sinα)²/(2g). Enter initial velocity, launch angle, and optional initial height. Supports metric (m/s) and imperial (ft/s) units.

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Calculation Parameters

m/s
°
m

Enter Parameters

Fill in the form on the left and click "Calculate"



What is Maximum Height of a Projectile?

The maximum height of a projectile is the highest vertical position it reaches during its trajectory. At this point, the vertical component of velocity becomes zero — the projectile has stopped rising and is about to begin falling back down.

Maximum Height Formula

The object reaches its maximum height when the vertical velocity equals zero (vy = 0). Setting the vertical velocity equation to zero gives the time to reach peak height:

Time to maximum height:
th = v₀ · sin(α) / g

Substituting this time into the vertical displacement equation yields the maximum height formula:

Maximum height:
hmax = h₀ + (v₀ · sin α)² / (2g)

Where:

  • h₀ — initial height above the ground (m or ft)
  • v₀ — initial launch speed (m/s or ft/s)
  • α — launch angle above the horizontal (degrees)
  • g — gravitational acceleration (9.807 m/s² or 32.174 ft/s²)

How to Find the Maximum Height of a Projectile

  1. Select your unit system: metric (m/s, m) or imperial (ft/s, ft).
  2. Enter the initial velocity v₀ — the speed at which the object is launched.
  3. Enter the launch angle α — the angle above the horizontal (between 0° and 90°).
  4. Optionally enter the initial height h₀ if the object is launched from an elevated position.
  5. Click Calculate to instantly find the maximum height, time to peak, and velocity components.

Examples

Example 1 — Ground-level launch (metric)

A ball is kicked at 20 m/s at an angle of 45° from the ground (h₀ = 0):

vy0 = 20 × sin(45°) ≈ 14.14 m/s
th = 14.14 / 9.807 ≈ 1.44 s
hmax = 0 + (14.14)² / (2 × 9.807) ≈ 10.20 m

Example 2 — Elevated launch (imperial)

A baseball is thrown at 90 ft/s at 30° from a height of 6 ft:

vy0 = 90 × sin(30°) = 45 ft/s
th = 45 / 32.174 ≈ 1.40 s
hmax = 6 + (45)² / (2 × 32.174) ≈ 37.43 ft

Frequently Asked Questions

What angle gives the maximum height?

A launch angle of 90° (straight up) gives the greatest maximum height for a given initial speed, since all of the velocity is directed vertically. However, a 90° launch gives zero horizontal range. For the best combination of range and height, 45° is the classic choice.

Does air resistance affect the maximum height?

Yes — in real-world scenarios, air resistance reduces the maximum height compared to the theoretical formula. This calculator assumes ideal projectile motion (no air drag), which is a good approximation for dense, slow-moving objects over short distances.

Can the initial height be negative?

This calculator requires h₀ ≥ 0. If you need to model a launch below ground reference level, you can set your reference point accordingly so that h₀ = 0 corresponds to your launch point.

What is the relationship between maximum height and range?

Maximum height and horizontal range are both functions of the launch angle. As the angle increases from 0° to 90°, the maximum height increases, while the horizontal range peaks at 45° and then decreases. Use the Projectile Motion Calculator to find range, time of flight, and other quantities simultaneously.

Calculation History

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