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A super ball bounces by 2

3
of its initial bounce when
dropped on a hard surface. If it is dropped from a
height of β„Žπ‘π‘š.

29. After what bounce would the ball have
travelled a total distance of 2.74β„Žπ‘π‘š
A. 4
B. 5
C. 6
D. ∞

The total distance travelled by the ball after 𝑛 bounces can be calculated using the formula:

total distance = π‘›β„Ž + 2(π‘›βˆ’1)π‘˜β„Ž

where π‘˜=23 is the fraction of the initial height reached after each bounce.

So, we need to find the value of 𝑛 for which the total distance is 2.74β„Ž.

2.74β„Ž = π‘›β„Ž + 2(π‘›βˆ’1)π‘˜β„Ž

Simplifying and solving for 𝑛, we get:

𝑛 = 3.74/0.69 β‰ˆ 5.42

Therefore, the ball would have travelled a total distance of 2.74β„Ž after the 5th bounce.

Answer: B. 5

To determine the answer, we need to find the total distance traveled by the ball after each bounce.

Given that the ball bounces by 2/3 of its initial bounce height, we can calculate the distance traveled in each bounce as follows:

Distance after the first bounce = 2/3 * β„Žπ‘π‘š
Distance after the second bounce = 2/3 * (2/3 * β„Žπ‘π‘š)
Distance after the third bounce = 2/3 * (2/3 * (2/3 * β„Žπ‘π‘š))

We can see that for each successive bounce, the distance traveled is 2/3 of the previous distance traveled.

Now, let's calculate the total distance traveled after each bounce:

Total distance after the first bounce = Distance after the first bounce
Total distance after the second bounce = Distance after the first bounce + Distance after the second bounce
Total distance after the third bounce = Distance after the first bounce + Distance after the second bounce + Distance after the third bounce

Simplifying these equations, we have:

Total distance after the first bounce = 2/3 * β„Žπ‘π‘š
Total distance after the second bounce = 2/3 * β„Žπ‘π‘š + 2/3 * (2/3 * β„Žπ‘π‘š)
Total distance after the third bounce = 2/3 * β„Žπ‘π‘š + 2/3 * (2/3 * β„Žπ‘π‘š) + 2/3 * (2/3 * (2/3 * β„Žπ‘π‘š))

Now, let's calculate the total distance after each bounce:

Total distance after the first bounce = 2/3 * β„Žπ‘π‘š = (2/3)β„Žπ‘π‘š

Total distance after the second bounce = (2/3)β„Žπ‘π‘š + (2/3)(2/3)β„Žπ‘π‘š = (2/3)β„Žπ‘π‘š + (4/9)β„Žπ‘π‘š

Total distance after the third bounce = (2/3)β„Žπ‘π‘š + (4/9)β„Žπ‘π‘š + (8/27)β„Žπ‘π‘š = (2/3)β„Žπ‘π‘š + (4/9)β„Žπ‘π‘š + (8/27)β„Žπ‘π‘š

From the given information, we are looking for the bounce at which the total distance traveled is 2.74β„Žπ‘π‘š.

Comparing the total distances after each bounce to 2.74β„Žπ‘π‘š, we find that the total distance after the third bounce is the closest match.

Therefore, the answer is option C. 6