Given that,
The initial speed of a ball, u = 17.3 m/s
It is shot from the roof of a 52 m tall building, h = 52 m
It is required to find the distance from the base of the building the ball land. Let the distance be d.
Final velocity of the ball, v = 0 (when it lands)
We can use third equation of motion to find it.
[tex]v^2-u^2=2ad[/tex]
a = -g
[tex]-u^2=-2gd\\\\d=\dfrac{u^2}{2g}\\\\d=\dfrac{(17.3)^2}{2\times 9.8}\\\\d=15.26\ m[/tex]
So, it will land at a distance of 15.26 m.
