Suppose you are on a rooftop and you throw a ball downward to the ground with an

Question

Suppose you are on a rooftop and you throw a ball downward to the ground with an initial speed of v(0). You then throw an identical ball upward with the same initial speed v(0). After rising, this ball also falls to the ground. At the moment each strikes the ground, how do the speeds of the balls compare? Use the concept of conservation of mechanical energy in your reasoning. Ignore air resistance.

Explanation / Answer

down-ball> projected down with = +v ME initial = final KE + PE = KE1 + PE1 0.5 mv^2 + mgh = 0.5 m vg^2 + 0 vg^2 = v^2 + 2gh ----------------------------- (1) ================================== up-ball> projected down with = - v >>>>>>>> note ME initial = final KE + PE = KE1 + PE1 0.5 m(-v)^2 + mgh = 0.5 m vg1^2 + 0 vg1^2 = v^2 + 2gh ----------------------------- (2) 2 equations return same ground velocities because +v or - v do not alter KE of course, time of flight will be different, but energy conservation is independent of time.

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