We considered a body of mass m = 0.4 kg which falls from rest for a time of 0.2 seconds. It falls from point A to a point 0.196 m below. The lower point is called point B. Suppose this is the ground. Potential energy is measured from the ground. At B the object's velocity is v = 1.96 m/s.

These results were found as follows: With the acceleration of gravity equal to g = 9.8 m/s2, the velocity after 0.2 seconds is v = gt = (9.8)(0.2) = 1.96 m/s. The distance fallen is D = (1/2)gt2 = (1/2)(9.8)(0.2)2 = 0.196 m. Potential energy at A is given by PEA = mgD = (0.4)(9.8)(0.196) = 0.768 J. Kinetic energy at A, since the mass starts at rest is zero. Total energy at A is TEA = PEA + KEA = 0.768 + 0 = 0.768 J. At B, the potential energy is zero. The kinetic energy is KEB = (1/2)mv2 = (1/2)(0.4)(1.96)2 = 0.768 J. Total energy at B is TEB = 0 + 0.768 = 0.768 J.

Now consider a time t = 0.1 s after the mass is released from point A. It falls a distance y from A to point C. y is given by

Therefore, as shown in the figure, the height of point C above the ground is 0.196 - 0.049 = 0.147 m. Call this height z. The potential energy at C is

The velocity at point C is given by v = gt = (9.8)(0.1) = 0.98 m/s. Therefore the kinetic energy is

So the total energy at C is

We verify that the total energy is the same at points A, B, and C.