A prankster drops a water balloon from the top of a building on an unsuspecting person on the sidewalk below. If the balloon is traveling at when it strikes a person's head (above the ground), how tall is the building

Given:

Initial velocity, vo = 0 m/s (at rest)

Final velocity, vf = 22.5 m/s

Height of the person, h = 1.5 m

Using equations of motion,

vf^2 = vo^2 + 2aS

Where,

S = distance of the motion

22.5^2 = 0 + (2 * 9.81 * S)

S = 506.25/19.62

= 25.80 m

Height of the building, H = distance of motion + height of the person

= 25.8 + 1.5

= 27.3 m

H = 28.5m

Explanation:

v = 23.7 m/s

u = 0

distance travelled use the relation,

v^2 = u^2 + 2gs

s = v^2/2g = 23.7^2/2x9.8 = 27 m

The height of the building is,

H = s + 1.5 m = 27 + 1.5 = 28.5 m