66.0k views
4 votes
Suppose you drop a ball from a window 45 meters above the ground. The ball bounces to 65% of its previous height with each bounce. What is the total number of meters (rounded to the nearest tenth) the ball travels between the time it is dropped and the 10th bounce?

User Zrneely
by
6.8k points

2 Answers

0 votes

Final answer:

To find the total number of meters the ball travels between the time it is dropped and the 10th bounce, we need to sum up the distances of each bounce. Using the formula for the sum of a geometric series, we can calculate the total distance the ball travels.

Step-by-step explanation:

To find the total number of meters the ball travels between the time it is dropped and the 10th bounce, we need to sum up the distances of each bounce.

Starting with the initial height of 45 meters, the first bounce will reach 65% of that height, which is 29.25 meters.

The ball will then bounce back up to 65% of the previous height, which is 19.00625 meters. This process continues until the 10th bounce, where the height will be 65% of the previous height, which is approximately 1.50966 meters.

Summing up all these distances, we get a total of 45 + 29.25 + 19.00625 + ... + 1.50966 meters.

To find this sum, we can use the formula for the sum of a geometric series:

S = a(1 - r^n) / (1 - r)

where S is the sum of the series, a is the first term (45 meters), r is the common ratio (0.65), and n is the number of terms (10 bounces).

Plugging in these values, we can calculate the total distance the ball travels.

User Maxdow
by
7.4k points
4 votes
208.7 meters hope this helps
User Ted Petrou
by
7.2k points