Question

The height a ball bounces is less than the height of the previous bounce due to friction. The heights of the bounces form a geometric sequence. Suppose a ball is dropped from one meter and rebounds 95​% of the height of the previous bounce. What is the total distance traveled by the ball when it comes to​ rest? Does the problem give you enough information to solve the​ problem? How can you write the general term of the​ sequence? What formula should you use to calculate the total​ distance?

Answer 1

Answer:

Explanation:

Let the first height be h . second height .75h

third height .75h . fourth height .75²h

fifth height .75²h , sixthth height .75³ and so on

Total distance consists of two geometric series as follows

1 ) first series

h + .75h + .75²h + .75³h......

2 )  second series

.75h +.75²h +.75³h + .75⁴h .......

Sum of first series :

first term a = h , commom ratio r = .75

sum = a / (1 - r )

= h / 1 - .75

= h / .25

4h

sum of second series :--

first term a = .75 h , commom ratio r = .75

sum = a / (1 - r )

= .75h / 1 - .75

= .75h / .25

3h

Total of both the series

= 4h + 3h

= 7h .

h = 1 m

Total distance = 7 m