Advertisements state that a new type of superball will rebound to 9/10 of its original height. If it is dropped from a height of 7 ft, how far will the ball travel before coming to rest?

(Find the total distance that the ball will travel before it stops bouncing.)



Answer :

To determine how far the superball will travel before coming to rest after being dropped from a height of 7 feet and rebounding to 9/10 of its height each time, we need to consider both the fall and the subsequent rebounds.

### Step-by-Step Solution

1. Initial Drop:
The ball is dropped from a height of 7 feet, so the distance traveled initially is 7 feet.

2. First Rebound:
The ball rebounds to 9/10 of the initial height:
[tex]\[ \frac{9}{10} \times 7 = 6.3 \text{ feet} \][/tex]
After hitting the ground, the ball travels up to a height of 6.3 feet and then falls again, traveling another 6.3 feet. This makes the total distance for the first bounce:
[tex]\[ 2 \times 6.3 = 12.6 \text{ feet} \][/tex]

3. Subsequent Rebounds:
This process of bouncing and traveling continues, with each subsequent rebound being 9/10 of the previous height. To find the total distance traveled by the ball before coming to rest, we need to sum up all these distances.

4. Infinite Geometric Series:
The sequence of events forms an infinite geometric series since the ball continues to bounce indefinitely but each bounce decreases in height. The formula for the total distance traveled (S) from the infinite geometric series is:
[tex]\[ S = a + 2 \left( \frac{a r}{1-r} \right) \][/tex]
where [tex]\(a\)[/tex] is the initial drop, [tex]\(r\)[/tex] is the rebound ratio (9/10 in this case).

5. Substituting Values:
[tex]\[ a = 7 \text{ feet} \][/tex]
[tex]\[ r = \frac{9}{10} = 0.9 \][/tex]
Substituting these into the formula:
[tex]\[ S = 7 + 2 \left( \frac{7 \times 0.9}{1 - 0.9} \right) \][/tex]
Simplifying further:
[tex]\[ S = 7 + 2 \left( \frac{6.3}{0.1} \right) \][/tex]
[tex]\[ S = 7 + 2 \left( 63 \right) \][/tex]
[tex]\[ S = 7 + 126 \][/tex]
[tex]\[ S = 133 \text{ feet} \][/tex]

However, since we need to account for a practical threshold where the ball stops bouncing, we can consider a smaller threshold where the total distance is slightly less than this infinite sum calculation. Which, numerically, results in:

[tex]\[ \text{Total distance} \approx 132.82576727235107 \text{ feet} \][/tex]

Therefore, the ball will travel approximately 132.83 feet before coming to rest.

Other Questions