show that the maximum height reached by a particle in projectile motion is ... ymax = Vo^2sin^2(theta)o/ 2g

Answer 1

Say, the initial velocity is Vo with an angle theta above the ground. The horizontal speed is Vo*cos(theta) and the vertical speed is Vo*sin(theta). Since the deccelaration is -g, the time to take to reach ymax is:
Vo*sin(theta)/g
Therefore: ymax = 0.5*g*t^2
= 0.5*g*{Vo*sin(theta)/g}^2
= Vo^2sin^2(theta)/(2g)