Answer:
a) The equation for ball´s velocity is V₁ = (F*t)/m
b) The ball velocity after collision is 7.87 m/s
c) The heigh from which the ball is dropped is 3.16 m
Explanation:
Given data:
m = mass of golf ball = 0.19 kg
t = interaction time = 0.13 s
F = force = 11.5 N
a) Due the collision is eslastic, the change of momentum is:
ΔV*m = F*t
V₁ - V₀ = (F*t)/m
V₁ = ((F*t)/m) + V₀
Where V₀ = 0
V₁ = (F*t)/m
b) Replacing:
[tex]V_{1} =\frac{11.5*0.13}{0.19} =7.87m/s[/tex]
c) According conservation of kinetic energy:
(Ek - Ep)initial = (Ek - Ep)final
0 + mgh = (1/2)mV₁² + 0
Clearing h:
[tex]h=\frac{1}{2}\frac{V_{1}^{2} }{g} =\frac{7.87^{2} }{2*9.8} =3.16m[/tex]
