This question involves the concepts of the equations of motion both in the horizontal motion and the vertical motion.
The rock hits "30.91 m" away from the base of the building.
First, we will calculate the time taken by the rock to reach the ground by applying the second equation of motion in vertical motion:
[tex]h=v_it+\frac{1}{2}gt^2[/tex]
where,
h = height of the building = 30 m
vi = initial vertical speed = 0 m/s
t = time = ?
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]30\ m = (0\ m/s)(t)+\frac{1}{2}(9.81\ m/s^2)(t)^2\\\\t=\sqrt{\frac{(2)(30\ m)}{9.81\ m/s^2}}\\\\[/tex]
t = 2.47 s
Now, we will calculate the horizontal distance by analyzing horizontal motion. Ignoring the air friction, the horizontal motion will be uniform:
[tex]s=vt[/tex]
where,
s = horizontal distance = ?
v = horizontal speed = 12.5 m/s
Therefore,
[tex]s=(12.5\ m/s)(2.47\ s)[/tex]
s = 30.91 m
Learn more about equations of motion here:
brainly.com/question/5955789?referrer=searchResults
