A ball is dropped from a height of 256 ft. With each bounce, the ball reaches a height that is half the height of the previous bounce. After which bounce will the ball rebound to a maximum height of 1 ft?
To find the number of bounces before the ball rebounds to a maximum height of 1 ft, we can set up a geometric sequence.
Let's denote the initial height of the ball as h_0 = 256 ft. We know that with each bounce, the ball reaches a height that is half the height of the previous bounce. So, the heights of the bounces can be calculated as follows:
h_0 = 256 ft (initial height)
h_1 = h_0 / 2 = 256 ft / 2 = 128 ft (height after the first bounce)
h_2 = h_1 / 2 = 128 ft / 2 = 64 ft (height after the second bounce)
h_3 = h_2 / 2 = 64 ft / 2 = 32 ft (height after the third bounce)
...
We can observe that the heights form a geometric sequence with a common ratio of 1/2. The general formula for the height after n bounces can be written as:
h_n = h_0 * (1/2)^n
We need to find the value of n when h_n (maximum height) is equal to 1 ft. Let's set up the equation:
1 = 256 * (1/2)^n
To solve for n, we can take the logarithm base 2 of both sides of the equation:
log2(1) = log2(256 * (1/2)^n)
0 = log2(256) + log2((1/2)^n)
0 = log2(256) + n * log2(1/2)
Since log2(1) = 0, the equation simplifies to:
0 = log2(256) + n * (-1)
Now, isolate n by moving the logarithm term to the other side:
n = -log2(256)
Using logarithm properties, we can rewrite -log2(256):
n = log2(1/256)
Finally, we can evaluate the logarithm:
n = log2(1) - log2(256)
n = 0 - 8
n = -8
However, since the number of bounces cannot be negative, we must ignore the negative solution.
Therefore, the ball will rebound to a maximum height of 1 ft after the 8th bounce.