a student drop a rubber ball from a height of 8 feet . Each time the ball hits the ground, it bounces to 75% of its previous height.
a/ how far does the ball travel between tne second and third bounces .
b/ write an infinite series to model the total distance tralveled by the ball, excluding the distance traveled before the firstbounce
a/ Well, let's see. The ball bounces back to 75% of its previous height, so after the first bounce, it reaches a height of (8 * 0.75) feet. After the second bounce, it reaches a height of ((8 * 0.75) * 0.75) feet. So, to find the distance traveled between the second and third bounces, we need to calculate the difference between the height at the second bounce and the height at the third bounce. And that would be ((8 * 0.75) * 0.75) - (((8 * 0.75) * 0.75) * 0.75) feet. Simple, right?
b/ Now, for an infinite series to model the total distance traveled by the ball, excluding the distance traveled before the first bounce... Hmm, I'd say it's a pretty "bouncy" situation. Let's call the height of the initial drop H. So, the first bounce would be at H, the second would be at H * 0.75, the third would be at (H * 0.75) * 0.75, and so on. So, we can define the series as H + H * 0.75 + (H * 0.75) * 0.75 + ...
But since we're talking about an infinite series here, it might take a while for the ball to stop bouncing. It's like the Energizer Bunny of balls!
To find the distance traveled by the ball between the second and third bounces, we need to calculate the distance covered by the ball during the second and third bounces.
a/ Distance covered during the second bounce:
The ball hits the ground and bounces back to 75% of its previous height, which is 8 ft * 0.75 = 6 ft.
Therefore, the distance covered during the second bounce is twice the height, which is 6 ft * 2 = 12 ft.
b/ Distance covered during the third bounce:
After the second bounce, the ball reaches a height of 6 ft * 0.75 = 4.5 ft.
Again, the distance covered during the third bounce is twice the height, which is 4.5 ft * 2 = 9 ft.
Therefore, the ball travels a total distance of 12 ft + 9 ft = 21 ft between the second and third bounces.
To write an infinite series to model the total distance traveled by the ball, excluding the distance traveled before the first bounce, we can use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
where:
S = sum of the series
a = the first term of the series (distance covered during the first bounce)
r = the common ratio (percentage of the height after each bounce)
In this case, the first term a = 2 * 8 ft (distance covered during the first bounce) = 16 ft, and the common ratio r = 0.75.
So, the infinite series to model the total distance traveled by the ball (excluding the distance traveled before the first bounce) is:
S = 16 ft / (1 - 0.75)
To solve these two questions, we need to understand the pattern of the ball's bounces and apply the concept of geometric series. Let's break it down step by step:
Step 1: Understanding the bounce pattern
- The ball is dropped from a height of 8 feet.
- After the first bounce, it reaches 75% (0.75) of its previous height.
- From the second bounce onwards, it follows the same pattern - each subsequent bounce reaches 75% of the previous height.
Step 2: Calculating the distance traveled between the second and third bounces
To find the distance traveled between the second and third bounces, we need to calculate the sum of the distances traveled during each bounce.
- The distance traveled during the second bounce is 0.75 times the distance traveled during the first bounce.
- The distance traveled during the third bounce is 0.75 times the distance traveled during the second bounce.
So, let's assume the distance traveled during the first bounce is "d" feet.
- Distance traveled during the second bounce = 0.75 * d
- Distance traveled during the third bounce = 0.75 * (0.75 * d) = 0.5625 * d
Thus, the distance traveled between the second and third bounces is the sum of these two distances:
- Distance = (0.75 * d) + (0.5625 * d) = 1.3125 * d
Therefore, the ball travels 1.3125 times the distance traveled during the first bounce between the second and third bounces.
Step 3: Writing the infinite series to model the total distance traveled
The total distance traveled by the ball can be represented by an infinite series because the number of bounces is unlimited.
Let's assume the distance traveled during the first bounce is "d" feet.
- Distance traveled during the second bounce = 0.75 * d
- Distance traveled during the third bounce = 0.75 * (0.75 * d) = 0.5625 * d
- Distance traveled during the fourth bounce = 0.75 * (0.5625 * d) = 0.421875 * d
- Distance traveled during the fifth bounce = 0.75 * (0.421875 * d) = 0.31640625 * d
- And so on...
We can see that the distance traveled during each bounce forms a geometric series with a common ratio of 0.75.
Therefore, the total distance traveled by the ball (excluding the distance traveled before the first bounce) can be represented as an infinite geometric series:
Distance = d + 0.75 * d + 0.5625 * d + 0.421875 * d + 0.31640625 * d + ...
This infinite series can be simplified using the formula for the sum of an infinite geometric series:
Distance = d / (1 - 0.75)
Simplifying further:
Distance = 4d
So, the total distance traveled by the ball (excluding the distance before the first bounce) can be expressed as 4 times the distance traveled during the first bounce.
we have to count the up and the down of each bounce
So from the moment of the first bounce:
distance = 2(.75)(8) + 2(.75^2)(8) + 2(.75^3)(8) + ...
= 16(.75) + 16(.75^2) + 16(.75^3) + ...
so we have a GS, with
a = 16 ,r = .75
distance between 3rd and 2nd
= 16(.75^2) - 16(.75^3) = 2.25 ft
total distance = a/(1-r) = 16/(1-.75) = 64 ft