Finding an expression for projectile's max height

A projectile is launched with speed v0 and angle (theta). Derive an expression for the projectile's maximum height, h

thanks for your help!

Hmax = Vy²/2g (constant acceleration with time eliminated)

Hmax = v0²*sin²Θ/(2g)

v_vertical = v_v = vo * sin(theta)
g = 9.81 or you can leave it as g (which we will)
h_max happens when v_final = 0 m/s

Basic formula vf*2 = vi^2 + 2*a*d
0 = vi^2 * sin^2(theta) + 2*g*h_max
h_max = - vi^2 * sin^2(theta) / ( 2* g)

The reason h_max is minus is because it is going in the opposite direction to the gravitation constant.

50 ft