Answer:
 a) 3.414 seconds
 b) 57.12 m
 c) 34.14 m
Explanation:
You want the vertical and horizontal distance traveled by a ball thrown horizontally from the top of a building at 10 m/s, given it falls half the height in the last second.
a) Fall time
The equation for the height of the ball can be written ...
 h(t) = -4.9t² +h₀
where t is the fall time in seconds, and hâ‚€ is the initial height in meters.
We want to find t such that ...
 h(t) = 0
 h(t -1) = h₀/2
The first of these equations tells us ...
 4.9t² = h₀
And the second of these equations tells us ...
 2·4.9(t -1)² = h₀ = 4.9t²
Dividing by 4.9 and subtracting t², we have ...
 t² -4t +2 = 0
 t² -4t +4 = 2 . . . . add 2 to complete the square
 (t -2)² = 2 . . . . . . . . . . write as a square
 t = 2 +√2 ≈ 3.414 . . . . . . . square root, add 2
The total fall time is about 3.414 seconds.
b) Height
 h₀ = 4.9·(2+√2)² = 4.9(6 +4√2) ≈ 57.12 . . . . meters
The height of the building is about 57.12 meters.
c) Horizontal
The horizontal distance traveled is the product of speed and time:
 (10 m/s)(3.414 s) = 34.14 m
The ball hits about 34.14 meters from the base of the building.
