Question

A ball is dropped from a height of 10 feet. After bouncing, it reaches a height that is 80% of the height from which it was dropped. This bouncing trend continues until the ball comes to rest. Show that the distance the ball falls on each drop follows a geometric sequence.

Answer 1

The distance the ball falls forms a geometric sequence because with each bounce, it reaches a height that is 80% of the height from which it was dropped, thus having a constant ratio of 0.8.

Step-by-step explanation:

The distance a ball falls after each bounce forms a geometric sequence because it follows a constant ratio. In this case, the ball is dropped from a height of 10 feet and bounces back to 80% of its previous height. This means the first drop is from 10 feet, the second drop (after the first bounce) is 80% of 10 feet, which is 8 feet, the third drop is 80% of 8 feet, which is 6.4 feet, and so on.

The general form for a geometric sequence can be expressed as a, a×r, a×r2, a×r3, ..., where a is the first term and r is the common ratio. For the bouncing ball, a = 10 feet and r = 0.8 (since 80% can be written as 0.8). This yields the sequence 10, 8, 6.4, 5.12, ... showcasing that each term is obtained by multiplying the previous term by the ratio of 0.8. As the ball continues to bounce, it covers less distance vertically after each bounce, following this geometric progression.