When “superballs” sprang upon the scene in the 1960’s, kids across the United States were amazed that these hard rubber balls could bounce to 90% of the height from which they were dropped. If a superball is dropped from a height of 2 m, how far does it travel by the time it hits the ground floor for the tenth time?

Question

A vivid experimental scene showing a glossy superball, characteristic of the 1960s, being dropped from a height of 2 meters onto a ground floor. The superball bounces repetitively, reaching slightly less height with each bounce. The trajectory should show the increasingly shorter bounces, up to the tenth bounce when the ball finally comes to rest on the floor. This rapid bouncing and the marvel of the rebounding superball should be captured with colorful, playful, and dynamic aesthetics but without any numerical or text explanations. The scene should be in a simple setting, perhaps an outdoor playground or a child's room.

When “superballs” sprang upon the scene in the 1960’s, kids across the United States were amazed that these hard rubber balls could bounce to 90% of the height from which they were dropped. If a superball is dropped from a height of 2 m, how far does it travel by the time it hits the ground floor for the tenth time?

Answer 1

set this up as a geometric series and then just apply the sum formula.

a = 2
r = 0.9

allow for round trips up and down

Answer 2

To determine how far the superball will travel by the time it hits the ground floor for the tenth time, we need to calculate the total distance it travels after each bounce.

The superball bounces to 90% of the height from which it was dropped. So, after the first bounce, it reaches a height of 2m * 0.9 = 1.8m.

Since the ball will keep bouncing, it will reach a height of 1.8m for the second bounce, then 1.8m * 0.9 = 1.62m for the third bounce, and so on.

We can see that the height after each bounce follows a geometric sequence with the initial term (a = 2) and the common ratio (r = 0.9).

To find the total distance traveled by the time it hits the ground floor for the tenth bounce, we need to sum up the distances traveled after each individual bounce.

The formula for the sum of the terms of a geometric sequence is given by: Sn = a * (1 - r^n) / (1 - r), where Sn is the sum of n terms, a is the initial term, r is the common ratio, and n is the number of terms.

Let's use this formula to calculate the total distance traveled by the superball after ten bounces:

n = 10 (number of terms)
a = 2 (initial term)
r = 0.9 (common ratio)

Sn = 2 * (1 - 0.9^10) / (1 - 0.9) = 2 * (1 - 0.387420489) / 0.1
Sn = 2 * (0.612579511) / 0.1
Sn ≈ 12.25159022

The total distance traveled by the superball by the time it hits the ground floor for the tenth time is approximately 12.25 meters.

Answer 3

To determine how far the superball travels by the time it hits the ground floor for the tenth time, we can set up a simple calculation.

First, let's figure out how far the superball travels each time it bounces. We know that it bounces to 90% of the height from which it is dropped, so if it is dropped from a height of 2m, it will bounce to 0.9 * 2m = 1.8m.

Now, we can calculate the total distance the superball travels on the way down to the ground floor for each bounce.

For the first drop, the distance traveled will be 2m. When it bounces back up, it will travel an additional 1.8m. So, the total distance covered after the first drop and bounce is 2m + 1.8m = 3.8m.

For the second drop and bounce, the distance traveled will be 1.8m (the height of the bounce), plus the distance traveled in the next drop and bounce. So, the total distance covered after the second drop and bounce is 1.8m + 3.8m = 5.6m.

We can continue this pattern for each drop and bounce, summing up the distances traveled each time until we reach the tenth drop.

To save time, we can use the formula for the sum of the first n terms of a geometric sequence:

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

Here, a represents the first term (2m), n represents the number of terms (in this case, 10 drops), and r represents the common ratio (0.9, because the superball bounces to 90% of its previous height).

Using the formula, we can calculate the total distance covered:

S = 2m * (1 - 0.9^10) / (1 - 0.9)

Simplifying this calculation:

S = 2m * (0.34867726...) / (0.1)

S ≈ 6.973545...m

Thus, by the time it hits the ground floor for the tenth time, the superball would have traveled approximately 6.97 meters.