Question

A 2.0-kg ball rolling along a flat surface starts up a hill . If the ball reaches a height of 0.63 m, what was it’s initial speed ?

Answer 1

PE = mph
= (2.0 kg) (9.8m/s^2)(0.63m)
=12.35 J

KE = 1/2mv^2
v^2= KE/1/2m
=12.35 J/1/2 (2.0kg)

v^2 = 12.35 m^2/s^2
v = 3.51 m/s