A cannonball is fired on flat ground at 420 m/s at a 53.0 degree angle. What maximum height does it reach?
The maximum height reached is given by
Hmax = v ^2 [ sin θ ]^2 / [2 g ] where v is the initial velocity, θ = 53° and g = 9.8 .....so we have
Hmax = [420]^2 * [ sin (53) ]^2 / [ 2 * 9.8 ] ≈ 5740.37 m
