Maximum Height of a Projectile

What this formula says

A projectile launched from the ground with speed $u$ at angle $\theta$ reaches a maximum height:

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

This is the highest point of its trajectory — the point where the projectile momentarily stops moving upward before beginning its descent.

The key insight

At the maximum height, the vertical component of velocity is zero.

The projectile still has horizontal velocity at this point — it is moving forward. But it has stopped moving upward. The vertical velocity has been reduced to zero by gravity.

This condition — $v_y = 0$ at the top — is what we use to find $H$.

Derivation

The vertical motion alone determines the maximum height. Horizontal motion is irrelevant here.

Initial vertical velocity: $u_y = u\sin\theta$

Acceleration in vertical direction: $-g$ (gravity, downward)

At maximum height $H$: final vertical velocity $v_y = 0$

Apply the third equation of motion to the vertical direction:

$v_y^2 = u_y^2 + 2(-g)H$

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

$2gH = u^2\sin^2\theta$

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

Why the third equation is the natural choice here

The third equation $v^2 = u^2 + 2as$ relates velocity, acceleration, and displacement — with no time. Since we are given the condition on velocity ($v_y = 0$) and asked for a displacement ($H$), the third equation is the direct route. We do not need to find the time to reach the top.

Alternatively, using the first equation to find time first, then the second to find height:

From $v_y = u_y - gt_{\text{top}}$, with $v_y = 0$:

$t_{\text{top}} = \frac{u\sin\theta}{g}$

Substituting into $s = u_y t - \frac{1}{2}gt^2$:

$H = u\sin\theta \cdot \frac{u\sin\theta}{g} - \frac{1}{2}g\left(\frac{u\sin\theta}{g}\right)^2$

$H = \frac{u^2\sin^2\theta}{g} - \frac{1}{2}g \cdot \frac{u^2\sin^2\theta}{g^2}$

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

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

Same result, more steps. The third equation is cleaner.

How $\theta$ affects maximum height

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

$\sin^2\theta$ increases from $0°$ to $90°$, so:

Relation between $H$ and $T$

The time to reach maximum height is $t_{\text{top}} = \frac{u\sin\theta}{g} = \frac{T}{2}$.

This confirms the symmetry: the trajectory is a perfect parabola, and the top is exactly at the halfway point in time.