Question 1090861
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?
<pre>
It falls 81 meters and hits the ground for the 1st time,
so its first upward bounce is two-thirds of 81m or 54m.

a<sub>1</sub> = 54, r = 2/3

We use geometric series formula for nth term 

a<sub>n</sub> = a<sub>1</sub>r<sup>n-1</sup>

a<sub>5</sub> = 54(2/3)<sup>5-1</sup> = 54(2/3)<sup>4</sup> = 54(16/81) = 32/3 m = {{{10&2/3}}} meters.

-------------</pre>what is the total distance it has covered the 6th time it touch the ground?<pre>The ball falls 6 times, but rises only 5 times.

We use the sum formula for geometric series:

{{{a[n]=(a[1](1-r^n))/(1-r)}}}

For the 6 falls, a<sub>1</sub> = 81m, r = 2/3

{{{a[6]=(81(1-(2/3)^6))/(1-2/3)=221&2/3}}} meters

For the 5 rises, a<sub>1</sub> = 54m, r = 2/3

{{{a[5]=(54(1-(2/3)^5))/(1-2/3)=140&2/3}}} meters

Total distance covered = total falls + total rises = {{{221&2/3+140&2/3=362&1/3}}} meters.

Edwin</pre></b>