A ball is dropped from $405$ meters and rebounds two-thirds the distance it falls each time it bounces. How many meters will the ball have traveled when it hits the ground the fourth time
To find the distance traveled by the ball when it hits the ground the fourth time, we can use the concept of a geometric sequence.
First, let's find the distance the ball travels during the first bounce. Since it rebounds two-thirds the distance it falls, it will travel $\frac{2}{3} \times 405$ meters.
Next, let's find the distance the ball travels during the second bounce. It will fall the same distance it did during the first bounce ($\frac{2}{3} \times 405$ meters), and then rebound two-thirds of that distance: $\frac{2}{3} \times \frac{2}{3} \times 405$ meters.
Following this pattern, the distance traveled during each bounce will be:
Bounce 1: $\frac{2}{3} \times 405$ meters
Bounce 2: $\left(\frac{2}{3}\right)^2 \times 405$ meters
Bounce 3: $\left(\frac{2}{3}\right)^3 \times 405$ meters
To find the distance traveled when the ball hits the ground the fourth time, we calculate the distance traveled during the first three bounces and add them together:
Distance traveled when hitting the ground the fourth time = $\frac{2}{3} \times 405 + \left(\frac{2}{3}\right)^2 \times 405 + \left(\frac{2}{3}\right)^3 \times 405$
Evaluating this expression, we get:
Distance traveled when hitting the ground the fourth time = $270 + 180 + 120 = 570$ meters
Therefore, the ball will have traveled $570$ meters when it hits the ground the fourth time.
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