Question

HELP...PLEASE

A ball that bounces up half the distance from which it is dropped is released three metres from the floor. If it continues bouncing forever, what is the total distance it travels (in m)?

Answer 1

The total distance a ball travels that bounces up half the distance from which it was dropped, with an initial drop of 3 meters, is calculated using the sum of an infinite geometric series. The sum is found to be 6 meters.

Step-by-step explanation:

The student's question involves calculating the total distance traveled by a ball that bounces indefinitely, where each bounce is half the height of the previous bounce. To answer this question, you can use the concept of geometric series because the distance the ball travels constitutes an infinite series with a common ratio of 1/2. The first bounce causes the ball to travel 3 meters down and then 1.5 meters up, for a total of 4.5 meters on the first bounce. Since the ball travels half the previous height with each bounce, the total distance followed by the ball can be expressed as a geometric series:

• First bounce (down + up): 3 m + 1.5 m = 4.5 m
• Second bounce (down + up): 1.5 m + 0.75 m = 2.25 m
• Third bounce (down + up): 0.75 m + 0.375 m = 1.125 m
• And so on...

The series is: S = 3 + 2(1.5) + 2(0.75) + 2(0.375) + ..., which can be simplified to S = 3 + 2(3/2 + 3/4 + 3/8 + ...). The sum of this infinite geometric series where r = 1/2 is S = a / (1 - r), where a is the first term of the series. In this case, a = 3 and r = 0.5, so the sum S is equal to:

S = 3 / (1 - 0.5) = 3 / 0.5 = 6 meters.

Hence, if the ball continues to bounce forever, the total distance it will travel is 6 meters.

Answer 2

Step-by-step explanation:

This will be a sum of two geometric sequences, one for distance falling and one for distance rising. The sum of a geometric sequence is expressed as s=a(1-r^n)/(1-r) where a=initial term, r=common ratio, and n= number of terms.

For the falling distance we have

f=3(1-.5^n)/(1-.5) as n goes to infinity it is just f=3/(.5)

For rising distance we have

r=1.5(1-.5^n)/(1-.5) and again as n goes to infinity it is r=1.5/(.5)

So the total distance travelled is

d=f+r

d=3/0.5+1.5/0.5

d=6+3

d=9 meters