posted by Lauren .
A cement block accidentally falls from rest from the ledge of a 51.7-m-high building. When the block is 14.9 m above the ground, a man, 2.05 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?
free fall from 51.7 to 14.9 meters - how fast is it going?
easy way is energy
(1/2) m v^2 = m g (51.7-14.9)
v^2 = 2 * 9.81 * 36.8
v = 26.7 m/s initial speed down at 14.9 height
how long to fall another (14.9 -2.05)= 12.85 meters?
increase in Ke = m g h
(1/2) m v^2 = (1/2)m(26.7)^2 + m g (12.85)
v^2 = 2* 26.7^2 + 2*9.81(12.85)
v^2 = 1677
v = 41 m/s when it hopefully misses
average speed between sighting and missing = .5(41 + 26.7) = 33.85 m/s
time to fall 12.85 m = 12.85/33.85 = .38 seconds (forget it)
Your formula is wrong, i tried it twice already.