Number of times the ball bounces back when dropped from a building of height h

I need help in calculating the math behind the below question.

A child plays with a ball on the nth floor of a big building the height of which is known . He lets out the ball. The ball rebounds for example to two-thirds of its height.

His mother looks out of a window that is 1.5 meters from the ground.

How many times will the mother see the ball either falling or bouncing in front of the window

Note
You will admit that the ball can only be seen if the height of the rebouncing ball is stricty greater than the window parameter.

Example:
h = 3, bounce = 0.66, window = 1.5, result is 3

What i have learnt is

I know that the ball falls from 27 feet and bounces back up 2/3 of that height, so I multiplied 27 times (2/3) and got 18 feet for the height of the first bounce.

Then, since it went up 18 feet, it fell from 18 feet
for the second bounce and bounced back up 2/3 of 18 feet. A table showing the four bounces:

  1. Bounce 1: (2/3) * 27 feet = 18 feet

  2. Bounce 2: (2/3) * 18 feet = 12 feet

  3. Bounce 3: (2/3) * 12 feet = 8 feet

I dono from this , how to reach solution ?

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An answer

Let’s start by defining some variables. Let $h$ be the height, $b$ be the percentage of the height which it bounces, and $v$ be the maximum possible height the ball is visible to the mother.

Every time the ball bounces, we take the height of the ball and multiply it by the bounce $b$. We are doing this repeated, and as you may recall repeated multiplication is exponentiation. So, for $n$ bounces, the height will be $hb^n$, or the height times the bounce factor $n$ times. Now, for the ball to be visible, the height of the bounce must be greater than the variable $v$. From this, we can construct an inequality $hb^n > v$. Now, we can solve for $n$ and round down.

Example: $h=3$, $b=2/3$, $v=1.5$ (as stated in your example)

From substitution, we get $3(2/3)^n > 1.5$. We can simplify this to get

$$(2/3)^n>0.5$$

$$log_{2/3}(2/3)^n > log_{2/3}(0.5)$$

$$n > 1.709$$

The ball will bounce $1$ time before it is out of visibility.

We can check this by doing your method

Bounce 1: $3 * (2/3) = 2$

Bounce 2: $2 * (2/3) = 4/3 < 1.5$

At bounce 2, the ball is no longer visible.

Formal Solved Equation

We can even make a formal equation for any $h, b, n, v$.

$$hb^n > v$$

$$b^n > v/h$$

$$log_bb^n > log_bv/h$$

$$n > log_bv/h$$

Which when rounded down gives

$$n = \lfloor log_bv/h \rfloor$$

for $\lfloor\rfloor$ is the floor function, which rounds down a number to an integer.