Problem Solution: 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	180	0	1800	1800	-1800
1	10	175	50	1750	1800	-1700
2	20	160	200	1600	1800	-1400
3	30	135	450	1350	1800	-900
4	40	100	800	1000	1800	-200
5	50	55	1250	550	1800	+700
6	60	0	1800	0	1800	+1800
					Total	-3500

(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	180	450	1800	2250	-1350
1	30	150	450	1500	1950	-1050
2	30	120	450	1200	1650	-750
3	30	90	450	900	1350	-450
4	30	60	450	600	1050	-150
5	30	30	450	300	750	+150
6	30	0	450	0	450	+450
					Total	-3150

(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	-5	180	12.5	1800.0	1812	-1788
1	6.67	179.17	22.2	1791.7	1814	-1769
2	18.33	166.67	168.1	1666.7	1835	-1499
3	30	142.5	450.0	1425.0	1875	-975
4	41.67	106.67	868.1	1066.7	1935	-199
5	53.33	59.17	1422.2	591.7	2014	+830
6	65	0	2112.5	0	2113	+2113
					Total	-3286