First I’ll show you this standard derivation using conservation of energy: Pi=Kf, mgh = 1/2 m v^2, V = sqrt(2gh) P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity. In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))= 5.477 m/s.
