**A super ball is dropped from a height off 2 m and ounces 90% of its original height on each bounce. 1. When it hits the ground for the eight time, how far has it traveled? 2. How high off the floor is the ball at the time of the eighth bounce?**

**Answer:**

Dropped from, a = 2m

1st bounce, a1 = 90% of 2 = 1.8m

2nd bounce, a2 = 90% of 1.8 = 1.62m

and so on.

this gives us the geometric progression

To find any term of a geometric sequence:

where “a=2” is the first term of the sequence,

r=0.9(from 90%) is the common ratio, n=8 is the number of the term to find.

we have,

an = a*r^(n-1)
a8th term = 2*[0.9^(8-1)]

= 2*[0.53]

=1.06 m far is the floor from ball during 8th bounce.

and

sum of the certain numbers in GP is given by

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

where Sn is Sum up to the nth term which is 8 here,

so

S8 = [2(1-0.9^8)] / (1-0.9)

= [2(1-0.477)] / 0.1

= [2*0.523]/0.1

= 10.46 m is the distance “x” covered by ball till 8th bounce.