Answer:
20.47 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{24^2-0^2}{2\times 9.81}\\\Rightarrow s=29.3577\ m[/tex]
Total height of the fall is 29.3577 m
Height the ball reached above the building is [tex]29.3577-8=21.3577\ m[/tex]
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 21.3577=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{21.3577\times 2}{9.81}}\\\Rightarrow t=2.0866\ s[/tex]
Time taken to reach the point from where the ball was thrown is 2.0866 s
This will also be the time it takes the ball to reach the maximum height
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow u=\dfrac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\dfrac{21.3577-\frac{1}{2}\times -9.81\times 2.0866^2}{2.0866}\\\Rightarrow u=20.47\ m/s[/tex]
The initial velocity with which the rock was thrown was 20.47 m/s
