A ball is dropped from rest from height h above the floor. At the instant the first ball hits the floor, a second identical ball is dropped also from rest and from height *H* above the floor. At what height above the floor will the two balls meet? (Assume zero air resistance, elastic collision with the floor, and dimensions of balls are negligible compared to *H*)

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Answer:

3*H*/4

Let’s set *t*=0 to be the instant when the 1^{st} ball has just rebounded, and the 2^{nd} ball just dropped.

Graph 1 and graph 2 (see below) are the speed-time graphs for the 1^{st} and 2^{nd} ball respectively. The gradient in both graphs correspond to g (the acceleration of free fall). The area-under-the-*v*–*t*-graph represents the displacement. So in (both graphs) the entire triangular area would correspond to *H*.

Stare at the graphs long enough (see below), and you should realize that the two balls will meet at time *T*/2, when ball 1 has risen a distance of 3*H*/4, and ball 2 has dropped a distance of *H*/4, and , for a total distance of *H*.