Question

If a rubber ball rebounds to 97% Of its height with each bounce, how many times will it bounce before rebounds to half of its original height?

Answer 1

If a quantity A is decreased to a x% of its initial value, then the new value will be:

A*(x%/100%).

Using this, we will see that the ball rebounds 23 times before rebounds to half of its original height.

Ok, let's assume that the original height of the rebound is H.

After one rebound, the new height will be H*(97%/100%) = H*0.97

After another rebound, the new height will be: (H*0.97)*(97%/100%) = H*(0.97)^2

You can already see the pattern here. This is an exponential decay, and the height of the n-th rebound can be written as:

f(n) = H*(0.97)^n

We want to find the value of n such that the height of that rebound is exactly half of the initial height, so we want to solve:

f(n) = H/2 = H*(0.97)^n

We can divide both sides by H to get:

1/2 = (0.97)^n

Now remember the natural logarithm property:

ln(x^k) = k*ln(x)

Then we can apply the natural logarithm to both sides to get:

ln(1/2) = ln((0.97)^n)

ln(1/2) = n* ln(0.97)

ln(1/2)/ln(0.97) = n = 22.76

Rounding to the next whole number, we have n = 23

This means that the ball will bounce 23 times before rebounds to half of its original height.

If you want to learn more, you can read:

brainly.com/question/19599469