Answer:
The building is 26.85m tall.
Explanation:
In order to solve this problem, we must start by drawing a diagram of the situation. (See attached picture).
We split the height of the building into three parts: [tex]y_{1}[/tex], the height of the window and [tex]y_2[/tex]
In order to find each of those, we need to start by finding the velocities of the ball in points A and B. We will analyze the trajectory of the ball when bouncing back to the top of the building. Let's start by finding the velocity of the ball in A:
We can use the following formula to determine the velocity of the ball in part A:
[tex]\Delta y=V_{A}t+\frac{1}{2}at^{2}[/tex]
which can be solved for [tex]V_{A}[/tex]
so we get:
[tex]V_{A}=\frac{\Delta y-\frac{1}{2}at^{2}}{t}[/tex]
and now we can substitute (remember the acceleration of gravity goes downward so we will consider it to be negative).
[tex]V_{A}=\frac{(1.30m)-\frac{1}{2}(-9.8m/s^{2})(0.115s)^{2}}{0.115s}[/tex]
which yields:
[tex]V_{A}=11.87m/s[/tex]
once we got the velocity at point A, we can now find the velocity at point B. We can do so by using the following formula:
[tex]a=\frac{V_{A}-V_{B}}{t}[/tex]
which can be solved for [tex]V_{B}[/tex] which yields:
[tex]V_{B}=V_{A}-at[/tex]
so we can substitute values now:
[tex]V_{B}=11.87m/s-(-9.8)(0.115s)[/tex]
Which yields:
[tex]V_{B}=13m/s[/tex]
Now that we have the velocities at A and B, we can use them to find the values of [tex]y_{1}[/tex] and [tex]y_{2}[/tex]
Let's start with [tex]y_{1}[/tex]
We can use the following formula to find it:
[tex]y_{1}=\frac{V_{f}^{2}-V_{A}^{2}}{2a}[/tex]
we know the final velocity of the rebound will be zero, so we can simplify our formula:
[tex]y_{1}=\frac{-V_{A}^{2}}{2a}[/tex]
so we can substitute now:
[tex]y_{1}=\frac{-(11.87m/s)^{2}}{2(-9.8m/s^{2})}[/tex]
which solves to:
[tex]y_{1}=7.19m[/tex]
Now we can proceed and find the value of [tex]y_{2}[/tex]
the value of [tex]y_{2}[/tex] can be found by using the following formula:
[tex]y_{2}=V_{B}t-\frac{1}{2}at^{2}[/tex]
in this case our t will be half of the tie spent below the bottom of the window, so:
[tex]t=\frac{2.04s}{2}=1.02s[/tex]
so now we can substitute all the values in the given formula:
[tex]y_{2}=(13m/s)(1.02s)-\frac{1}{2}(-9.8m/s^{2})(1.02s)^{2}[/tex]
which yields:
[tex]y_{2}=18.36m[/tex]
so now that we have all the values we need, we can go ahead and calculate the height of the building:
[tex]h=y_{1}+window+y_{2}[/tex]
when substituting we get:
[tex]h=7.19m+1.3m+18.36m[/tex]
So the answer is:
h=26.85m
The building is 26.85m tall.
