A cement block accidentally falls from rest from the ledge of a 51.5 -m-high building. When the block is 12.9 m above the ground, a man, 1.60 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?

V^2 = Vo^2 + 2g*d

V^2 = 0 + 19.6(51.5-12.9) = 756.56
V = 27.5 m/s.

d = Vo*t + 0.5g.t^2 = 12.9-1.6 = 11.3 m
d = 27.5*t + 4.9t^2 = 11.3
4.9t^2 + 27.5t - 11.3 = 0
Use Quadratic Formula and get:
t = 0.385 s.

To find the maximum time the man has to get out of the way, we need to calculate the time it takes for the block to fall from a height of 12.9 m.

We can use the equation of motion for free fall:

s = ut + (1/2)gt^2

where:
s = distance fallen (12.9 m)
u = initial velocity (0 m/s as the block is falling from rest)
g = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
t = time taken

Substituting the values into the equation, we get:

12.9 = (0)(t) + (1/2)(-9.8)(t^2)
12.9 = (-4.9)(t^2)

Rearranging the equation, we have:

t^2 = 12.9 / -4.9
t^2 ≈ -2.63

Since time cannot be negative, there is no real solution to this equation. Therefore, the block does not reach a height of 12.9 m before hitting the ground.

Consequently, the man has zero seconds to get out of the way.

To find the maximum time the man has to get out of the way, we need to determine the time it takes for the cement block to fall from a height of 12.9 m above the ground.

We can use the equation of motion for free fall:

h = ut + (1/2)gt²

Where:
h = height
u = initial velocity (in this case, 0 m/s since the block is falling from rest)
g = acceleration due to gravity (approximately 9.8 m/s²)
t = time

Rearranging the equation, we get:

t = √(2h/g)

Substituting the values, we have:

t = √(2 * 12.9 / 9.8)

Calculating this, we find:

t ≈ 1.43 seconds

Therefore, the man has approximately 1.43 seconds to get out of the way before the cement block reaches the height of 12.9 m above the ground.