Maximum height uses $v^{2}$ = $u^{2}$ + 2as with v = 0

Projectile Motion — AQA A-Level Physics

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

$H$: maximum height (m)
$u$: initial speed (m s⁻¹)
$θ$: angle of projection
$g$: accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻². due to gravity (m s⁻²)

Worked Example

A ball is thrown from point P with initial velocity 12 m s^-1 at 50 degrees to the horizontal. Find the maximum height.

Show Solution

Identify vertical motion variables

Apply $v^{2}$ = $u^{2}$ + 2as

$$0 = (12 \sin 50°)^2 - 2(9.81)H$$ $$H = \frac{(12 \sin 50°)^2}{2 \times 9.81} = \frac{(9.19)^2}{19.62} = 4.3 \text{ m}$$

Answer

$H = 4.3$ m

Maximum height uses \(v^{2}\) = \(u^{2}\) + 2as with v = 0