The end of a spring is pulled as far as it will go and then is released. On the first bounce back it extends 135 cm. On its second bounce back it extends 45 cm. On its third bounce back it extends 15 cm. How many times will the spring bounce back before it extends 5/9 cm?
I have no idea how to solve this problem. Could someone please show me the steps?
It's not the 6th bounce? Because when I put 6 into the equation I got 5/9
To solve this problem, let's analyze the pattern of the spring's movement.
We know that on the first bounce back, the spring extends 135 cm.
On the second bounce back, it extends 45 cm (which is one-third of the length of the first bounce), and on the third bounce back, it extends 15 cm (which is one-third of the length of the second bounce).
So, we can see that the length of each bounce back is decreasing by one-third each time.
To find out how many times the spring will bounce back before it extends 5/9 cm, we can set up an equation to solve:
135 * (1/3)^(n-1) = 5/9
Where "n" represents the number of bounces.
Let's solve this equation step-by-step:
Step 1: Rewrite 135 as a fraction with a denominator of 1.
135/1 * (1/3)^(n-1) = 5/9
Step 2: Cross-multiply to eliminate the fractions.
(135/1) * (9/5) = (1/3)^(n-1)
Step 3: Multiply the numerators and the denominators.
(135 * 9) / (1 * 5) = (1/3)^(n-1)
Step 4: Simplify.
1215/5 = (1/3)^(n-1)
Step 5: Solve (1/3)^(n-1) = 243.
Since 243 is 3^5, we can rewrite the equation as:
(1/3)^(n-1) = (1/3)^5
Step 6: Set the exponents equal to each other.
n-1 = 5
Step 7: Solve for n.
n = 5 + 1
n = 6
Therefore, the spring will bounce back 6 times before it extends 5/9 cm.
To solve this problem, we need to observe the pattern in the length of the spring with each bounce.
Let's denote the initial length of the spring as L.
On each bounce, the length of the spring decreases by a certain fraction. We can find this fraction by dividing the length of the previous bounce by the length of the current bounce.
In this case, the length decreases from 135 cm to 45 cm, which is a decrease of (135/45) = 3.
We can use this ratio, 3, to find the length of the next bounce. We divide 45 cm by 3 to get 15 cm.
Following this pattern, we can continue to divide the length of each bounce by 3 to find the length of subsequent bounces.
Using this approach, we can create a table to find the lengths of different bounces:
Bounce Number | Length of Bounce
-----------------------------
1 | 135 cm
2 | 45 cm
3 | 15 cm
4 | 5 cm
5 | 5/3 cm
6 | 5/9 cm
From the table, we can see that the spring will bounce back 6 times before it extends to 5/9 cm.
Therefore, the answer to the question is that the spring will bounce back 6 times before it extends to 5/9 cm.
Tn = 135/3^(n-1)
5/9 = 135/3^(n-1)
3^(n-1) = 135*9/5 = 81
since 3^4 = 81, n=5
So, on the 5th bounce it will extend 5/9 cm