Maximum Height of Projectile Formula: H = u² sin²θ / 2g

The maximum height of projectile formula is $H = \frac{u^2 \sin^2\theta}{2g}$ . It tells you how high above its launch point a projectile climbs when it leaves at speed u and angle θ to the horizontal. Only the upward part of the velocity, $u\sin\theta$ , lifts the projectile, which is why the sine appears squared.

For a ball thrown straight up, θ = 90° and the formula becomes $H = \frac{u^2}{2g}$ , a result you may already know from free fall. The rest of this page derives the formula, works three examples and shows how the height links to the range and the time of flight.

Quick facts

Formula
$H = \frac{u^2\sin^2\theta}{2g}$
Straight up
$H = \frac{u^2}{2g}$
Time to reach H
$\frac{u\sin\theta}{g}$
SI unit
metre (m)
Velocity at the top
$u\cos\theta$ , horizontal only
Link to range
$\frac{H}{R} = \frac{\tan\theta}{4}$

Deriving the maximum height of projectile formula

At the top of the arc the upward motion has just run out, so the vertical velocity is momentarily zero. Apply $v^2 = u^2 + 2as$ to the vertical motion only, with up as positive, a starting vertical velocity of $u\sin\theta$ and $a = -g$ :

$0 = (u\sin\theta)^2 - 2gH$

$H = \frac{u^2\sin^2\theta}{2g}$

The horizontal velocity plays no part in the climb. A ball thrown fast and flat barely rises, while the same speed aimed steeply goes much higher.

Three heights worked out

Example 1: a kicked ball

A volleyball is served at 24 m/s, aimed 30° upward. How high above the serving point does it rise?

$\sin 30^\circ = 0.5$ , so $\sin^2 30^\circ = 0.25$ .

$H = \frac{24^2 \times 0.25}{2 \times 9.8} = \frac{576 \times 0.25}{19.6} = \frac{144}{19.6} \approx 7.35\ \text{m}$

Example 2: a straight up throw

A coin is flicked vertically upward at 15 m/s. Find the greatest height above the hand.

$H = \frac{15^2}{2 \times 9.8} = \frac{225}{19.6} \approx 11.5\ \text{m}$

Example 3: a garden sprinkler

Water leaves a nozzle at 43.2 km/h at 60° to the horizontal. How high does the jet rise?

43.2 ÷ 3.6 = 12 m/s, and $\sin^2 60^\circ = 0.75$ .

$H = \frac{144 \times 0.75}{19.6} = \frac{108}{19.6} \approx 5.51\ \text{m}$

How maximum height relates to range and flight time

Dividing the height formula by the range of projectile formula gives a neat ratio:

$\frac{H}{R} = \frac{u^2\sin^2\theta / 2g}{u^2 \cdot 2\sin\theta\cos\theta / g} = \frac{\tan\theta}{4}$

Check it on Example 1: at 24 m/s and 30° the range on level ground would be 576 × 0.866 ÷ 9.8 ≈ 50.9 m, and 50.9 × tan 30° ÷ 4 ≈ 7.35 m, which matches. At 45°, tan θ = 1, so the height is a quarter of the range.

The peak is reached at half the time of flight, at t = u sin θ / g. In Example 1 that is 24 × 0.5 ÷ 9.8 ≈ 1.22 s after the serve.

What changes the height

Effect of each input on H
Change	Effect on H	Reason
Double u	Four times higher	H depends on u²
Raise θ toward 90°	Higher	sin θ grows
Stronger gravity	Lower	g is in the denominator
Heavier ball	No change	Mass does not appear

Lunar gravity is roughly a sixth of ours, so a throw that peaks at 2 m on Earth would peak near 12 m on the Moon. Every other input stays the same; only the denominator shrinks.

Common slips

Using sin θ instead of sin²θ. Square the sine after you find it.
Dropping the 2 in 2g. The denominator is 19.6 m/s² when g = 9.8 m/s².
Saying the speed is zero at the top. Only the vertical part is zero; the projectile still moves sideways at u cos θ.
Measuring from the ground when the launch was from a height. H is measured from the launch point.

All three projectile results grow from the vertical motion equations listed on the kinematics formulas hub, so learn those first and these follow.

Frequently asked questions

What is the formula for maximum height of a projectile?

H = u² sin²θ / 2g. It is measured from the launch point upward and treats the air as offering no drag.

Which launch angle gives the tallest arc?

At 90°, a vertical throw, where H = u² / 2g. Any lower angle sends part of the speed sideways.

What is the velocity of a projectile at maximum height?

Its vertical velocity is zero, but it still moves horizontally at u cos θ. Only for a vertical throw is the whole velocity zero at the top.

Does the mass of a projectile change its maximum height?

No. Without air resistance every object has the same downward acceleration, so mass cancels out.

How are maximum height and range related?

H / R = tan θ / 4. At a 45° launch, the maximum height is one quarter of the range.