A superball is dropped from rest from a height of 2.0m. It bounces repeatedly from the floor, as superballs are prone to do. After each bounce the ball dissipates some energy, so eventually it comes to rest. The following pattern is observed: After the 1st bounce, the ball returns to a maximum height that is 3/4 of its initial height. After the 2nd bounce, the ball returns to a maximum height that is 3/4 of its maximum height after the 1st; After the 3rd bounce, the ball returns to a maximum height that is 3/4 of its maximum height after the second, etc. In fact, for this particular ball, the maximum height is achieved after the nth bounce is found to be 3/4 of the maximum height achieved in the previous bounce. If this pattern is repeated, how many times will the ball bounce before coming to rest, and how long the process will take? (Neglect air friction.)

1st fall: 2m

nth bounce: 2*(3/4)^n

So, you have to decide what "at rest" means. At some point the friction will overcome the tendency to bounce, and the bounces will stop. In theory, the ball bounces forever, each time achieving less height.

So, once you decide how high the minimum acceptable bounce is, just solve for n.

For example, if the ball is considered at rest when the bounces are less than .01mm high, then

2(3/4)^n = .00001
(3/4)^n = .000005
n = (log .000005)/log(3/4) = 42.4 bounces

The time for the nth bounce is
t = √(2*(3/4)^n)/4.9) = √(2/4.9) √(3/4)^n

Figure the time for the 1st fall, then add up as many terms for the bounce times as you decide you need.

Take a peek at

http://www.wolframalpha.com/input/?i=sum+%E2%88%9A%282*%283%2F4%29^n%29%2F4.9%29

To solve this problem, we need to set up an equation for the maximum height achieved after each bounce. Let's assume the maximum height after the first bounce is H. According to the given information, the maximum height after the second bounce would be (3/4)H, after the third bounce would be (3/4)((3/4)H), and so on.

We can observe that the maximum height after each bounce forms a geometric sequence with a common ratio of (3/4). The general formula for the nth term of a geometric sequence is given by:

an = a * r^(n-1)

where an is the nth term, a is the first term, r is the common ratio, and n is the number of terms.

In this case, a = H and r = (3/4). We want to find the number of bounces before the ball comes to rest. This means we need to find the value of n for which the maximum height becomes zero.

Setting the nth term to zero, we have:

0 = H * (3/4)^(n-1)

Since H is non-zero, we can cancel it from both sides of the equation:

0 = (3/4)^(n-1)

Now, we need to solve for n. Taking the logarithm of both sides, we have:

ln(0) = ln[(3/4)^(n-1)]

The natural logarithm of zero is undefined, so this equation does not have a meaningful solution. Therefore, the ball will never come to rest.

As for the time it takes for the ball to complete each bounce, we can use the concept of vertical motion under gravity. The time taken to reach the maximum height and the time taken to fall back down are equal for each bounce.

The time taken for an object to reach its maximum height when dropped from rest is given by:

t = sqrt((2h)/g)

where h is the height and g is the acceleration due to gravity (approximately 9.8 m/s^2).

In this case, the height is 2.0 m. Plugging in the values, we have:

t = sqrt((2 * 2.0)/9.8)
t = sqrt(0.40816)
t ≈ 0.6394 seconds

Therefore, each bounce will take approximately 0.6394 seconds.

Note: It's important to keep in mind that the given pattern may not continue indefinitely in real-world scenarios due to factors such as imperfect elasticity of the ball and dissipative forces like air resistance.

To solve this problem, we need to determine how many times the ball will bounce before coming to rest and the total time it takes for the process.

Let's analyze the given pattern: After the first bounce, the ball reaches a maximum height that is 3/4 of its initial height. Let's call this height H1. After the second bounce, the ball reaches a maximum height that is 3/4 of H1, which we'll denote H2. And so on.

The height after each bounce follows a geometric sequence. The general term of this sequence can be written as:

H_n = (3/4)^(n-1) * H1

Where H_n is the maximum height after the nth bounce.

To determine when the ball comes to rest, we need to find the condition where the height becomes negligible. In other words, when the maximum height is close to zero, we can say the ball has come to rest.

Let's set up an equation for the condition:

H_n = (3/4)^(n-1) * H1 = 0

Since H1 is not zero, we can divide both sides of the equation by H1:

(3/4)^(n-1) = 0

To find the value of n, we can take the logarithm of both sides of the equation:

log((3/4)^(n-1)) = log(0)

Using logarithm properties, we can bring down the exponent:

(n-1) * log(3/4) = log(0)

Now, we need to analyze the right side. Logarithms are undefined for zero, so the equation does not have a solution. This means the maximum height will never reach zero, and the ball will keep bouncing indefinitely.

However, it's important to note that due to energy dissipation, the ball's maximum height will become infinitesimally small after a large number of bounces, and we can consider it effectively at rest.

To determine the time it takes for this process, we need to consider the time it takes for each bounce. Let's assume the time it takes for each bounce to be constant and denoted as T.

The time taken to reach the maximum height after each bounce, starting from the initial drop, would be:

T + T + T + T + ... + T (n times) = nT

Since the height of the bounces decreases geometrically, the time taken for each bounce decreases by the same ratio of 3/4.

To find the total time, we need to sum the time taken for each bounce until the ball effectively comes to rest. However, since the number of bounces is infinite, we cannot directly sum them.

However, we can consider the sum of the geometric series as the limit of a convergent series. The sum of an infinite geometric series is given by the formula:

Sum = a / (1 - r)

Where 'a' is the first term and 'r' is the common ratio. In this case, the first term 'a' would be T, and the common ratio 'r' would be 3/4.

Therefore, the total time it takes for the ball to effectively come to rest can be expressed as:

Total time = lim(n → ∞) (nT / (1 - 3/4))

Simplifying the expression:

Total time = T / (1 - 3/4)

Total time = 4T

So the time it takes for the process to complete is 4 times the time it takes for one bounce.

In conclusion, the ball will bounce indefinitely but effectively come to rest due to energy dissipation. The total time it takes for this process is four times the time it takes for one bounce.