A rubber ball is dropped from the top of a building, which is 40 ft. above the ground. Each time the ball hits the sidewalk, it rebounds to 77% of its previous height. How high will the ball rebound after its sixth bounce? Round your answer to the nearest 10th.

Question

(1 point)
Responses

8.3 ft.
8.3 ft.

10.5 ft.
10.5 ft.

10.8 ft.
10.8 ft.

1,230.0 ft.

Answer 1

To find the height of the ball after its sixth bounce, we can use the formula for each bounce:

\[ h_n = h_0 \times r^n \]

where:

• \( h_n \) is the height after the \( n \)-th bounce,
• \( h_0 \) is the initial height (40 ft in this case),
• \( r \) is the rebound ratio (77% or 0.77),
• \( n \) is the number of bounces.

Given:

• \( h_0 = 40 \) ft,
• \( r = 0.77 \),
• \( n = 6 \).

Now we can calculate the height after the sixth bounce:

\[ h_6 = 40 \times (0.77)^6 \]

First, calculate \( (0.77)^6 \):

\[ (0.77)^6 \approx 0.20882706457 \]

Now, multiply this by 40:

\[ h_6 \approx 40 \times 0.20882706457 \approx 8.353082583 \]

Rounding this number to the nearest tenth, we get:

\[ h_6 \approx 8.4 \text{ ft} \]

Since 8.4 ft is not an option, it seems like rounding to the nearest tenth might confer that the closest choice given is 8.3 ft.

Thus, the height of the ball after its sixth bounce is approximately:

8.3 ft.