Answer:
vf = 4.01 m/s
Explanation:
According to the law of conservation of energy:
[tex]Kinetic\ Energy\ Lost\ by\ Ball = Potential\ Energy\ Gained\ by\ the\ Ball \\\frac{1}{2}m(v_f^2-v_i^2) = mgh\\\\ v_f^2-v_i^2 = 2gh\\v_f^2 = 2gh + v_i^2\\[/tex]
where,
vf = final speed of ball at top of the ramp = ?
vi = initial speed of ball = 2.62 m/s
g = acceleration due to gravity = 9.81 m/s²
h = height = 0.47 m
Therefore,
[tex]v_f^2 = (2)(9.81\ m/s^2)(0.47\ m)+(2.62\ m/s)^2\\v_f = \sqrt{9.2214\ m^2/s^2+6.8644\ m^2/s^2}\\[/tex]
vf = 4.01 m/s
