Answer:
Height = 53.361 m
Explanation:
There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12
From the given information we make the following summary
[tex]u_{1}[/tex] = 0m/s
[tex]t_{1}[/tex] = t
[tex]u_{2}[/tex] = 43.12m/s
[tex]t_{2}[/tex] = (t-2.2)s
The distance by the first balloon is
[tex]D = u_{1} t_{1} + \frac{1}{2} at_{1}^2[/tex]
where
a = 9.8m/s2
Inputting the values
[tex]D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2[/tex]
The distance traveled by the second balloon
[tex]D = u_{2} t_{2} + \frac{1}{2} at_{2}^2[/tex]
Inputting the values
[tex]D = (43.12)(t-2.2) + \frac{1}{2} (9.8)(t-2.2)^2[/tex]
simplifying
[tex]D = 4.9t^2 + 21.56t -71.148[/tex]
Substituting D of the first balloon into the D of the second balloon and solving
[tex]4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s[/tex]
Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment
[tex]D = 4.9(3.3)^2\\ D = 53.361 m[/tex]
