Question

Evan drops a ball vertically from a height of 150ft. The peak height after each bounce is half the previous height. How many feet does the ball travel from the time he drops the ball until it reaches the peak height after the 6th bounce? Show all steps used to find the solution.(This is a homework question I am stuck on and is from the unit Geometric Series and Applications)

Answer 1

Let's solve this problem graphically first.

The situation described on the problem is shown below:

In the diagram the yellow lines represent the distance travelled downwards and the red lines the distance travelled upwards. We notice that we start at a height of 150 ft an each bounce the distance the ball go up is half the previous distance. If we add the distances for each line we have:

`150+2(75)+2((75)/(2))+2((75)/(4))+2((75)/(8))+2((75)/(16))+(75)/(32)=(14175)/(32)`

Therefore, the distance travelled by the ball from the time it is dropped until it reaches the peak height after the 6th bounce is 14175/32 ft. (This is approximately 442.97 ft)

Now, let's solve the problem algebraically. To do this we will analyse the distance travelled downward and distance travelled upward separately.

The sequence the yellow lines follows is: 150, 75, 75/2, .... From it we notice that we have a geometric series that starts at 150 and have a common ratio r=1/2. The partial sum of a geometric series is given by:

`S_n=(a(1-r^n))/(1-r)`

In this case we want until the sixth term, then n=6. Plugging the values we know we have:

`\begin{gathered} S_6=\frac{150\lbrack1-((1)/(2))\placeholder{⬚}^6\rbrack}{1-(1)/(2)} \\ S_6=(150(1-(1)/(64)))/((1)/(2)) \\ S_6=(150((63)/(64)))/((1)/(2)) \\ S_6=((4725)/(32))/((1)/(2)) \\ S_6=(4725)/(16) \end{gathered}`

From this we conclude that the distance travelled by the ball downwards is 4725/16 ft.

The sequence of the red lines is: 75, 75/2, 75/4, .... Once again, we have a geometric series with common ration 1/2 but in this case the initial value is 75. Then we have:

`\begin{gathered} S_6=\frac{75\lbrack1-((1)/(2))\placeholder{⬚}^6\rbrack}{1-(1)/(2)} \\ S_6=(75(1-(1)/(64)))/((1)/(2)) \\ S_6=(75((63)/(64)))/((1)/(2)) \\ S_6=((4725)/(64))/((1)/(2)) \\ S_6=(4725)/(32) \end{gathered}`

Hence, the distance travelled by the ball upwards is 4725/32 ft.

Adding the downward and upward distances we have:

`(4725)/(16)+(4725)/(32)=(14175)/(32)`

Therefore, the distance travelled by the ball is 14175/32 ft.

Note: The answers from the two methods we used are the same, as they should.