Question

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.

30. What is the total distance the ball convers
before it finally rests
A. 243−1ℎ
B. 243h
C. 3−1ℎ
D. 3h

Answer 1

Since the ball bounces by 2/3 of its previous height, the total distance covered by the ball after its n-th bounce can be written as:

h + (2/3)h + (2/3)^2h + ... + (2/3)^(n-1)h

This is a geometric series with first term h and common ratio 2/3. The sum of a geometric series is given by:

Sum = (first term)/(1 - common ratio)

Plugging in the values, we get:

Sum = h/(1 - 2/3) = 3h

So the total distance covered by the ball before it finally rests is 3h. Therefore, the answer is D. 3h.