A special rubber ball is dropped from the top of a building
81m high. if after the falling it bounces back to two-thirds
of the previous bouncing height, then how high will it bounce
the 5th time?
It falls 81 meters and hits the ground for the 1st time,
so its first upward bounce is two-thirds of 81m or 54m.
a1 = 54, r = 2/3
We use geometric series formula for nth term
an = a1rn-1
a5 = 54(2/3)5-1 = 54(2/3)4 = 54(16/81) = 32/3 m = meters.
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what is the total distance it has covered the 6th time it touch the ground?The ball falls 6 times, but rises only 5 times.
We use the sum formula for geometric series:
For the 6 falls, a1 = 81m, r = 2/3
meters
For the 5 rises, a1 = 54m, r = 2/3
meters
Total distance covered = total falls + total rises = meters.
Edwin