Question

Evan drops a ball vertically from a height of 150 ft. 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. ​

Answer 1

442.96875 ft.

Explanation:

Original Height (h) : 150 ft

Bounce 1 (h₁) : h/2 = 150/2 = 75 ft

Bounce 2 (h₂) : h₁/2 = 75/2 = 37.5 ft

Bounce 3 (h₃) : h₂/2 = 37.5/2 = 18.75 ft

Bounce 4 (h₄) : h₃/2 = 18.75/2 = 9.375 ft

Bounce 5 (h₅) : h₄/2 = 9.375/2 = 4.6875 ft

Bounce 6 (h₆) : h₅/2 = 4.6875/2 = 2.34375 ft

Finding the sum of bounces 1 - 5 :

• 75 + 37.5 + 18.75 + 9.375 + 4.6875
• 112.5 + 18.75 + 9.375 + 4.6875
• 131.25 + 9.375 + 4.6875
• 140.625 + 4.6875
• 145.3125 ft

Distance traveled :

• Original Height + 2 x Sum of Bounces (1 - 5) + Height of 6th bounce
• 150 + 2 x 145.3125 + 2.34375
• 150 + 290.625 + 2.34375
• 440.625 + 2.34375
• 442.96875 ft.

Answer 2

442.96875 ft

Explanation:

We can model the height of each bounce as a geometric series.

General form of a geometric sequence:
`a_n=ar^(n-1)`

(where
`a` is the initial term and
`r` is the common ratio)

If we let n be the number of bounces, then the initial term has to be:

`a_1=\frac12(150)=75`

since this is the height of the ball after the first bounce.

The common ratio r has already been given to us, as we have been told that the height of the bounce halves after each bounce.

`\implies r=\frac12=0.5`

Therefore, the geometric series to model the peak height of each bounce is:

`a_n=75(0.5)^(n-1)`

(where
`a_n` is the height and n is the number of bounces)

Therefore, to calculate the total number of feet the ball travels until it reaches the peak height of the 6th bounce:

Total = 150 + (2 × sum of 1, 2, 3, 4 & 5th bounces) + 6th bounce

To find the sum, use the geometric sum formula:

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

Therefore, the sum of the bounces 1 through 5 is:

`\implies S_5=(75(1-0.5^5))/(1-0.5)=145.3125`

Height of the 6th bounce:

`\implies a_6=75(0.5)^(6-1)=2.34375`

`\begin{aligned}\textsf{Total distance} & =150+2\left((75(1-0.5^5))/(1-0.5)\right)+75(0.5)^(6-1)\\ & = 150 + 2(145.3125)+2.34375\\ & = 150 + 290.625+2.34375\\ & =442.96875\: \sf ft \end{aligned}`