The maximum height of projectile formula is ______.

Step 1: Formula used

The maximum height of the object is the highest vertical distance from the horizontal plane along its trajectory.

Use the third equation of motion $v^2=u^2-2gs$

Where $v$ is the final velocity, $u$ is the initial velocity, $g$ is the acceleration due to gravity ($9.80m{s}^{-2}$), $s$ is the maximum vertical distance.

Step 2: Find the maximum height of the projectile

The final velocity is zero $(v=0)$

When a body is launched in projectile motion making an angle $\theta$ with the horizontal, its initial velocity has both horizontal and vertical components.

The vertical component of the body describes the influence of velocity in displacing the component vertically. The vertical component is sine of the angle.

So, the initial velocity in the y-direction is $u\sin \theta$, since the body is moving with an initial velocity of $u$ at an angle $\theta$ from the ground (Since height is being discussed here, we will take the vertical component).

The displacement in the y-direction $\left(s\right)$ will be the maximum height $\left(H\right)$ achieved by the projectile.

Using these values,

$$\begin{gathered} 0=u^2-2gs \\ \Rightarrow 0={\left(u\sin \theta \right)}^2-2gs \\ \Rightarrow 2gs={\left(u\sin \theta \right)}^2 \\ \Rightarrow s=\frac{{\left(u\sin \theta \right)}^2}{2g} \end{gathered}$$

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

Thus, the maximum height of the projectile formula is, $H=\frac{u^2{\sin }^2\theta }{2g}$.