力學問題~~~~~~~~help!!

By the law of conservation of energy
Loss of potential energy = Gain of potential energy
mgh = 1/2 mv^2
gh = 1/2 v^2
(10)(50 X 0.01) = 1/2 v^2
v^2 = 10
v = 3.16 ms^-1

泥膠在相撞前一刻的速度為3.16 ms^-1.

泥膠的PE=泥膠的KE
mgh=0.5mv^2
(0.08)(10)(0.5)=0.5(0.08)v^2
v =3.16ms-1