Answer :
Sure, let's go through each part of the question step-by-step.
### Part A: Height after the 3rd bounce
To find the height after the 3rd bounce, we can use the rebound ratio. The rebound ratio tells us the proportion of the previous height that the ball reaches after each bounce. The rebound ratio in this case is 75%, or 0.75.
If the initial height is [tex]\( h_0 = 16 \)[/tex] feet, then the height after each bounce reduces by a factor of 0.75.
[tex]\[ h_n = h_0 \times (rebound\ ratio)^n \][/tex]
For the 3rd bounce ([tex]\( n = 3 \)[/tex]):
[tex]\[ h_3 = 16 \times (0.75)^3 \][/tex]
Calculate [tex]\( (0.75)^3 \)[/tex]:
[tex]\[ (0.75)^3 = 0.75 \times 0.75 \times 0.75 = 0.421875 \][/tex]
Now multiply by the initial height:
[tex]\[ h_3 = 16 \times 0.421875 = 6.75 \text{ feet} \][/tex]
Therefore, the height after the 3rd bounce is 6.75 feet.
### Part B: Height after the 17th bounce
To find the height after the 17th bounce, we use the same formula with [tex]\( n = 17 \)[/tex]:
[tex]\[ h_{17} = 16 \times (0.75)^{17} \][/tex]
Calculate [tex]\( (0.75)^{17} \)[/tex]:
[tex]\[ (0.75)^{17} \approx 0.010019329 \][/tex]
Now multiply by the initial height:
[tex]\[ h_{17} = 16 \times 0.010019329 \approx 0.160309264 \text{ feet} \][/tex]
Therefore, the height after the 17th bounce is approximately 0.1603 feet.
### Part C: Total distance traveled after the 10th bounce
To find the total distance traveled after the 10th bounce, let's break it down:
1. The ball travels the initial drop distance.
2. It then rebounds up to a certain height and falls back down again repeatedly.
3. Each time it rebounds up and down counts as traveling twice the current rebound height.
Let [tex]\( h_0 = 16 \)[/tex] feet be the initial height.
The total distance, [tex]\( D \)[/tex], is:
[tex]\[ D = h_0 + 2 \sum_{i=1}^{10} h_0 \times (rebound\ ratio)^i \][/tex]
First term:
[tex]\[ D = 16 + 2 \sum_{i=1}^{10} 16 \times (0.75)^i \][/tex]
Factor out the constants:
[tex]\[ D = 16 + 32 \sum_{i=1}^{10} (0.75)^i \][/tex]
Now calculate [tex]\( \sum_{i=1}^{10} (0.75)^i \)[/tex]:
This is a geometric series where the sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of a geometric series is given by:
[tex]\[ S_n = a \frac{1-r^n}{1-r} \][/tex]
In our case, [tex]\( a = 0.75 \)[/tex], [tex]\( r = 0.75 \)[/tex], and [tex]\( n = 10 \)[/tex]:
[tex]\[ S_{10} = 0.75 \frac{1-(0.75)^{10}}{1-0.75} \][/tex]
First, calculate [tex]\( (0.75)^{10} \)[/tex]:
[tex]\[ (0.75)^{10} \approx 0.05631351 \][/tex]
Then:
[tex]\[ S_{10} = 0.75 \frac{1-0.05631351}{0.25} = 0.75 \frac{0.94368649}{0.25} \][/tex]
[tex]\[ S_{10} = 0.75 \times 3.77474596 \approx 2.83105947 \][/tex]
Now we can calculate the total distance:
[tex]\[ D = 16 + 32 \times 2.83105947 = 16 + 90.594 \][/tex]
[tex]\[ D = 106.594 \text{ feet} \][/tex]
Therefore, the total distance the ball will travel after the 10th bounce is approximately 106.594 feet.
### Part A: Height after the 3rd bounce
To find the height after the 3rd bounce, we can use the rebound ratio. The rebound ratio tells us the proportion of the previous height that the ball reaches after each bounce. The rebound ratio in this case is 75%, or 0.75.
If the initial height is [tex]\( h_0 = 16 \)[/tex] feet, then the height after each bounce reduces by a factor of 0.75.
[tex]\[ h_n = h_0 \times (rebound\ ratio)^n \][/tex]
For the 3rd bounce ([tex]\( n = 3 \)[/tex]):
[tex]\[ h_3 = 16 \times (0.75)^3 \][/tex]
Calculate [tex]\( (0.75)^3 \)[/tex]:
[tex]\[ (0.75)^3 = 0.75 \times 0.75 \times 0.75 = 0.421875 \][/tex]
Now multiply by the initial height:
[tex]\[ h_3 = 16 \times 0.421875 = 6.75 \text{ feet} \][/tex]
Therefore, the height after the 3rd bounce is 6.75 feet.
### Part B: Height after the 17th bounce
To find the height after the 17th bounce, we use the same formula with [tex]\( n = 17 \)[/tex]:
[tex]\[ h_{17} = 16 \times (0.75)^{17} \][/tex]
Calculate [tex]\( (0.75)^{17} \)[/tex]:
[tex]\[ (0.75)^{17} \approx 0.010019329 \][/tex]
Now multiply by the initial height:
[tex]\[ h_{17} = 16 \times 0.010019329 \approx 0.160309264 \text{ feet} \][/tex]
Therefore, the height after the 17th bounce is approximately 0.1603 feet.
### Part C: Total distance traveled after the 10th bounce
To find the total distance traveled after the 10th bounce, let's break it down:
1. The ball travels the initial drop distance.
2. It then rebounds up to a certain height and falls back down again repeatedly.
3. Each time it rebounds up and down counts as traveling twice the current rebound height.
Let [tex]\( h_0 = 16 \)[/tex] feet be the initial height.
The total distance, [tex]\( D \)[/tex], is:
[tex]\[ D = h_0 + 2 \sum_{i=1}^{10} h_0 \times (rebound\ ratio)^i \][/tex]
First term:
[tex]\[ D = 16 + 2 \sum_{i=1}^{10} 16 \times (0.75)^i \][/tex]
Factor out the constants:
[tex]\[ D = 16 + 32 \sum_{i=1}^{10} (0.75)^i \][/tex]
Now calculate [tex]\( \sum_{i=1}^{10} (0.75)^i \)[/tex]:
This is a geometric series where the sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of a geometric series is given by:
[tex]\[ S_n = a \frac{1-r^n}{1-r} \][/tex]
In our case, [tex]\( a = 0.75 \)[/tex], [tex]\( r = 0.75 \)[/tex], and [tex]\( n = 10 \)[/tex]:
[tex]\[ S_{10} = 0.75 \frac{1-(0.75)^{10}}{1-0.75} \][/tex]
First, calculate [tex]\( (0.75)^{10} \)[/tex]:
[tex]\[ (0.75)^{10} \approx 0.05631351 \][/tex]
Then:
[tex]\[ S_{10} = 0.75 \frac{1-0.05631351}{0.25} = 0.75 \frac{0.94368649}{0.25} \][/tex]
[tex]\[ S_{10} = 0.75 \times 3.77474596 \approx 2.83105947 \][/tex]
Now we can calculate the total distance:
[tex]\[ D = 16 + 32 \times 2.83105947 = 16 + 90.594 \][/tex]
[tex]\[ D = 106.594 \text{ feet} \][/tex]
Therefore, the total distance the ball will travel after the 10th bounce is approximately 106.594 feet.
