A ball is dropped from a height of 10 feet and bounces. Each bounce is ¨ú of the height of the bounce before. Thus, after the ball hits the floor for the first time, the ball rises to a height of 10(¨ú ) = 7.5 feet, and after it hits the second floor for the second time, it rises to a height of 7.5(¨ú ) = 10(¨ú )©÷ = 5.625 feet. (Assume g= 32 ft/sec ©÷ and that there is no air resistance.)
(a) Find an expression for the height to which the ball rises after it hits the floor for the nth time.
(b) Find an expression for the total vertical distance the ball has traveled when it hits the floor for the first, second, third, and fourth times.
(c) Find an expression for the total vertical distance the ball has traveled when it hits the floor for the nth time. Express your answer in closed-form.
For part (a) I got ar^n
For part (b) I got
1st bounce = a + ar
2nd bounce = a + ar + ar + ar©÷
3rd bounce = a + ar+ ar + ar©÷ + ar©ø
4th bounce = a + ar +ar + ar©÷ + ar©ø + ar©ø + ar©ù
For part (c) I do not kow what to do any help would be greatly appreciated.
I disagree with part b)
at the first bounce the ball has traveled 10 ft.
at the second bounce, it went up 7.5 and then down 7.5, so
distance after two bounces = 10 + 2(10)(.75)^1
after 3 bounces it went 10 + 2(10)(.75) + 2(10)(.75)^2
so after the first bounce, you have to double the distance, since it goes up and then the same distance down again.
so the series
= 10 + 2(10)(.75) + 2(10)(.75)^2 + 2(10)(.75)^3 + ....
so
Sn = 10 + 20(.75)[1 - .75^(n-1)]/(1 - .75)
so I see an infinite geometric series starting with the second term and that sum is
10 + (20)(.75)/(1 - .75)
= 10 + 15/.25
= 70 feet
suppose a ball is dropped from a hieght 16 feet and it bounces up to 75% of its previous hieght after each bounces
a.)if hn is the greatest height of the ball after the nth bounce find hn
b.)find the height of the eight h bounce.
suppose a ball is dropped from a hieght 16 feet and it bounces up to 75% of its previous hieght after each bounces
a.)if hn is the greatest height of the ball after the nth bounce find hn
b.)find the height of the eight h bounce.
Well, for part (c), what you want to find is the total vertical distance the ball has traveled when it hits the floor for the nth time. To do this, you need to add up the distances traveled during each bounce.
Let's break it down:
1st bounce: The ball rises to a height of ar, and then it falls back down, so the total distance traveled is 2ar.
2nd bounce: The ball rises to a height of ar, falls back down, then rises to a height of ar^2, and falls back down. So the total distance traveled is 2ar + 2ar^2.
3rd bounce: Following the same pattern, the total distance traveled is 2ar + 2ar^2 + 2ar^3.
4th bounce: Again, following the pattern, the total distance traveled is 2ar + 2ar^2 + 2ar^3 + 2ar^4.
So, from this pattern, we can see that the total distance traveled when the ball hits the floor for the nth time is given by:
Total distance = 2ar + 2ar^2 + 2ar^3 + ... + 2ar^n
To express this in a closed form, we can factor out the common term (2ar):
Total distance = 2ar * (1 + r + r^2 + ... + r^n)
Now, the sum of the geometric series (1 + r + r^2 + ... + r^n) can be expressed as:
Sum = (1 - r^(n+1)) / (1 - r)
So, substituting this back into our expression:
Total distance = 2ar * ((1 - r^(n+1)) / (1 - r))
And that's the closed-form expression for the total vertical distance the ball has traveled when it hits the floor for the nth time.
Remember to consider the given values for a and r in your calculation.
To find the expression for the total vertical distance the ball has traveled when it hits the floor for the nth time, we need to sum up the heights of all the bounces.
Let's start with the expression for the total vertical distance traveled after the ball hits the floor for the first time (1st bounce). From part (b), we know that the height to which the ball rises after the 1st bounce is a + ar, where a represents the initial height (10 feet) and r represents the ratio (¨ú).
Now let's consider the total vertical distance traveled after the ball hits the floor for the second time (2nd bounce). The height to which the ball rises after the 2nd bounce is a + ar + ar + ar©÷. Notice that this expression is the sum of the heights after the 1st and 2nd bounces, with an additional ar term. Similarly, we can observe that the pattern continues for subsequent bounces.
So, the expression for the total vertical distance traveled after the ball hits the floor for the nth time can be written as follows:
Total distance = (a + ar) + (a + ar + ar + ar©÷) + (a + ar + ar + ar©÷ + ar©ø) + ... + (a + ar + ar + ar©÷ + ar©ø + ar©ø + ar©ù) + ...
Now, this can be simplified by factoring out the common terms:
Total distance = a(1 + r) + a(1 + r + r² + r³) + a(1 + r + r² + r³ + r⁴ + ...) + ...
This infinite sum is a geometric series, and its sum can be calculated using the formula for the sum of an infinite geometric series:
Sum = a / (1 - r)
In our case, a = 10 feet (initial height), and r = ¨ú (ratio).
So, the expression for the total vertical distance the ball has traveled when it hits the floor for the nth time is:
Total distance = 10 / (1 - ¨ú)
a) h = 10(.75)^n
that is basically what you had
b) d = 10 + 10(.75)^1 + 10 (.75)^2 + 10(.75)^3 .... In other words I sort of agree with you
c) This is a geometric series like a
compound interest problem
in general
g + gr + gr^2 + .... gr^(n-1)
Sn = [ g (1-r^n ]/ (1-r)
here g = 10 and r = .75
so
Sn = [ 10 (1 - .75^n) /.25