# How do you determine the maximum height of a projectile?

Jul 21, 2015

You use the fact that the vertical component of the initial velocity is zero at maximum height.

#### Explanation:

When you launch a projectile at an angle $\theta$ from the horizontal, the initial velocity of the projectile will have a vertical and a horizontal component.

$\left\{\begin{matrix}{v}_{0 x} = {v}_{0} \cdot \cos \left(\theta\right) \\ {v}_{0 y} = {v}_{0} \cdot \sin \left(\theta\right)\end{matrix}\right.$

In order to determine the maximum height reached by the projectile during its flight, you need to take a look at the vertical component of its motion.

Vertically, the motion of the projectile is affected by gravity. This means that at maximum height, the vertical component of the initial speed will be zero.

The projectile will decelerate on its way to maximum height, come to a complete stop at maximum height, then starts its free fall descent towards the ground.

If you use the vertical component of its initial speed, you can write

underbrace(v_"h max"^2)_(color(blue)("=0")) = v_text(0y)^2 - 2 * g * h_"max"

This is equivalent to

${v}_{0 y}^{2} = 2 \cdot g \cdot {h}_{\text{max}}$

The maximum height reached by the projectile will thus be

${h}_{\text{max}} = {v}_{0 y}^{2} / \left(2 \cdot g\right) = {\left({v}_{0} \cdot \sin \left(\theta\right)\right)}^{2} / \left(2 \cdot g\right)$

${h}_{\text{max}} = \textcolor{g r e e n}{\frac{{v}_{0}^{2} \cdot {\sin}^{2} \left(\theta\right)}{2 \cdot g}}$