# What does $\Delta y_{max} = \frac{v_0^2\sin^2(\theta)}{2y}$ mean, and how does one calculate it?

I am learning about projectiles in my physics course and was recently introduced to this equation: $$\Delta y_{max} = \frac{v_0^2\sin^2(\theta)}{2y}$$ I am having a hard time making sense of it. How does one utilize it? When are you supposed to use it? How do you put it in the calculator? I ask because I don't know how to input sin^2, and I only see Sin and Sin inverse within the calculator.

Thank you

The equation is used to find the maximum height of an object undergoing projectile motion. Note it depends only on the initial vertical velocity, angle of launch and gravitational acceleration. Also $$\sin^2(x) = (\sin x)^2$$.
\begin{align} 0 & = \totald{\bracks{v_{0}\sin\pars{\theta}t - gt^{2}/2}}{t} \implies t = {v_{0}\sin\pars{\theta} \over g} \\[5mm] & v_{0}\sin\pars{\theta}t - {1 \over 2}\,gt^{2} = v_{0}\sin\pars{\theta}\, {v_{0}\sin\pars{\theta} \over g} - {1 \over 2}\,g\bracks{v_{0}\sin\pars{\theta} \over g}^{2} = \bbx{\color{#44f}{v_{0}^{2}\sin\pars{\theta} \over 2g}} \\ & \end{align}