Answer:
74.529 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.8 m/s²
For first ball
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow s=4.9t^2\ m[/tex]
For second ball
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=50.96\times (t-2.6)+\frac{1}{2}\times 9.8\times (t-2.6)^2\\\Rightarrow s=25.48t-99.372+4.9t^2[/tex]
As the displacement is equal
[tex]25.48t-99.372+4.9t^2=4.9t^2\\\Rightarrow t=\frac{99.372}{25.48}\\\Rightarrow t=3.9\ s[/tex]
[tex]s=4.9\times 3.9^2=74.529\ m[/tex]
So, height of the building is 74.529 m
