What is the max height reached in m?

Question

A projectile is launched with a speed of 37 m/s at an angle of 59° above the horizontal. What is the maximum height reached by the projectile during its flight?

Answer 1

Once you break down the velocity into X and Y components, you can treat it like i projectile that you fire straight into the air.

Vy=Cos[59]*37 m/s = 31.71 m/s

Vf^2 - Vi^2 = 2*g*Xf

Xf = (Vf^2 - Vi^2)/2*g = (31.71^2 - 0)/2*9.8 = 51.31 m

Answer 2

sin(59)*37=initial vertical velocity.=31.7m/s
.5m*(31.7)^2=m*g*h
.5*(31.7)^2/g=h=51.31meters