A bag of sand dropped by a would-be assassin from the roof of a building just misses Tough Tony, a gangster 2-m tall. The missile traverses the height of Tough Tony in 0.22 s, landing with a thud at his feet. How high was the building? Ignore friction.
The bag was traveling at average speed 2/.22 = 9.1 m/s when it passed him.
That speed is reached in t = 9.1/g = 0.93 seconds
The distance it fell in that time is
the height H = (g/2)t^2 = 4.2 m
That average speed is reached when the bag has travelled about half his body length. The building is actually 1 meter higher, or 5.2 m.
The method I used is approximate. For a more exact answer,
(Time to fall H) - (Time to fall H-2) = 0.22 s
sqrt(2H/g) - sqrt[2(H-2)/g] = 0.22 s
Solve for H
You will get about the same answer.