To determine the total distance traveled by the ball dropped from an initial height of 10 feet and bouncing up to [tex]\(\frac{2}{3}\)[/tex] of its previous height, let's track the path of the ball step-by-step until it completes the 30th bounce.
Here's the detailed procedure:
### Step 1: Initial Drop
The ball is initially dropped from a height of 10 feet. As soon as it hits the ground, it bounces back. This initial drop contributes 10 feet to the total distance.
### Step 2: Bouncing Sequence
When the ball hits the ground and bounces back, the height it reaches is given by:
[tex]\[ 10 \times \left(\frac{2}{3}\right) \][/tex]
From this height, it falls back to the ground, thus contributing twice that height to the total travel distance for each bounce (once going up, and once coming back down).
For each bounce:
- The height after each bounce can be described by [tex]\(\left(\frac{2}{3}\right)^n \times 10\)[/tex] where [tex]\(n\)[/tex] is the number of the bounce.
- Each return journey contributes twice the height of the bounce.
### Step 3: Summing the Distances
For a total of 30 bounces, the sequence for distances will look like:
1. Initial drop: 10 feet.
2. First bounce:
[tex]\[ 2 \times \left( 10 \times \frac{2}{3} \right) = 2 \times \left( \frac{20}{3} \right) \][/tex]
[tex]\[ \text{Distance during first bounce} = 2 \times \left( \frac{20}{3} \right) \][/tex]
3. Second bounce:
[tex]\[ 2 \times 10 \times \left( \frac{2}{3} \right)^2 \][/tex]
4. Continue similarly for each bounce [tex]\( n \)[/tex]:
[tex]\[ 2 \times 10 \times \left( \frac{2}{3} \right)^n \][/tex]
Summing these up, the total distance traveled can be computed from:
[tex]\[ 10 + 2 \sum_{n=1}^{29} 10 \left(\frac{2}{3}\right)^n + \text{final bounce back to height} \][/tex]
Thus
[tex]\[ \text{Total Distance} = 10 + 2 \times 10 \left(\frac{2}{3} + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^3 + \cdots \left(\frac{2}{3}\right)^{29}\right) + 10 \left(\frac{2}{3}\right)^{30} \][/tex]
### Step 4: Geometric Series
[tex]\[ S = \left(\frac{2}{3} + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^3 + \cdots \left(\frac{2}{3}\right)^{29}\right) \][/tex]
The sum [tex]\( S \)[/tex] of a geometric series is given by:
[tex]\[ S = a \frac{1 - r^n}{1 - r} \][/tex]
where [tex]\( a \)[/tex] is the first term and [tex]\( r \)[/tex] is the common ratio.
Here, [tex]\( a = \frac{2}{3} \)[/tex] and [tex]\( r = \frac{2}{3} \)[/tex].
So,
[tex]\[ S_{29} = \frac{\frac{2}{3}(1 - \left(\frac{2}{3}\right)^{29})}{1 - \frac{2}{3}} = \frac{\frac{2}{3}(1 - \left(\frac{2}{3}\right)^{29})}{\frac{1}{3}} = 2 \left(1 - \left(\frac{2}{3}\right)^{29}\right) \][/tex]
Then, the total distance [tex]\( D \)[/tex]:
[tex]\[ D = 10 + 20 \left(1 - \left(\frac{2}{3}\right)^{29}\right) + 10 \left(\frac{2}{3}\right)^{30} \][/tex]
Given this, summing all components we'll have:
[tex]\[ D \approx 49.999739245247454 \text { feet}\][/tex]
Thus, the total distance traveled by the ball is approximately:
[tex]\[ 49.999739245247454 \text{ feet} \][/tex]
