The total distance travelled by the ball after 𝑛 bounces can be calculated using the formula:
total distance = 𝑛ℎ + 2(𝑛−1)𝑘ℎ
where 𝑘=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 2.74ℎ.
2.74ℎ = 𝑛ℎ + 2(𝑛−1)𝑘ℎ
Simplifying and solving for 𝑛, we get:
𝑛 = 3.74/0.69 ≈ 5.42
Therefore, the ball would have travelled a total distance of 2.74ℎ after the 5th bounce.
Answer: B. 5
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 ℎ𝑐𝑚.
29. After what bounce would the ball have
travelled a total distance of 2.74ℎ𝑐𝑚
A. 4
B. 5
C. 6
D. ∞
1 answer
1 answer