A 14 kg cannon ball is fired from a cannon with muzzle speed of 1319 m/s at an angle of 19.1 ◦ with the horizontal. The acceleration of gravity is 9.8 m/s 2. Use conservation of mechanical energy to find the maximum height reached by first ball. Answer in units of m

hf = 9494.338 m

Explanation:

m = 14 Kg

vi = 1319 m/s

hi = 0

∅ = 19.1º

viy = vi*Sin ∅ = (1319 m/s) * Sin 19.1º = 431.6 m/s

vfy = 0

We can apply

Ei = Ef

Ki + Ui = Kf + Uf

⇒ 0.5*m*viy² + m*g*0 = 0.5*m*0² + m*g*hf

⇒ hf = (0.5*viy²) / (g)

⇒ hf = (0.5 * (431.6 m/s) ²) / (9.81 m/s²)

⇒ hf = 9494.338 m