Respuesta :
Answer:
442.96875 ft.
Step-by-step 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:
442.96875 ft
Step-by-step explanation:
We can model the height of each bounce as a geometric series.
General form of a geometric sequence: [tex]a_n=ar^{n-1}[/tex]
(where [tex]a[/tex] is the initial term and [tex]r[/tex] is the common ratio)
If we let n be the number of bounces, then the initial term has to be:
[tex]a_1=\dfrac12(150)=75[/tex]
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.
[tex]\implies r=\dfrac12=0.5[/tex]
Therefore, the geometric series to model the peak height of each bounce is:
[tex]a_n=75(0.5)^{n-1}[/tex]
(where [tex]a_n[/tex] 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:
[tex]S_n=\dfrac{a(1-r^n)}{1-r}[/tex]
Therefore, the sum of the bounces 1 through 5 is:
[tex]\implies S_5=\dfrac{75(1-0.5^5)}{1-0.5}=145.3125[/tex]
Height of the 6th bounce:
[tex]\implies a_6=75(0.5)^{6-1}=2.34375[/tex]
[tex]\begin{aligned}\textsf{Total distance} & =150+2\left(\dfrac{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}[/tex]
