Problem: A one kilogram object, initially at rest, falls from a height 180 meters above the earth. Approximating the acceleration of gravity as g =10 m/s², the object will strike the earth in 6 seconds.

(a) Compute the instantaneous height h and the speed v at one second intervals over the physical trajectory. From this compute the instantaneous values for the kinetic and potential energies, and their sum and difference. Total the differences for the seven intervals.

t	v=gt	h=180-gt^2/2	KE=mv^2/2	PE=mgh	KE+PE	KE-PE
0	0
1	10
2	20
3	30
4	40
5	50
6	60
					Total

(b) Repeat the calculation using a non-physical path in which the object travels the distance 180 meters in 6 seconds at the constant speed v=30 m/s.

t	v=30	h=180-vt	KE=mv^2/2	PE=mgh	KE+PE	KE-PE
0	30
1	30
2	30
3	30
4	30
5	30
6	30
					Total

( c) Determine speed of fall as an object thrown up at 5 m/s, which falls with an acceleration of a=(7/6)g = 70/6 m/s². Thus

v = -5 + (70/6)t ; h = 180 + 5t – (35/6)t²

t	v	h	KE=mv^2/2	PE=mgh	KE+PE	KE-PE
0
1
2
3
4
5
6
					Total