A cement block accidentally falls from rest from the ledge of a 51.6 -m-high building. When the block is 15.7 m above the ground, a man, 1.90 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?
To find the maximum time the man has to get out of the way, we need to find the time it takes for the block to fall from a height of 15.7 m to the ground.
We can use the equations of motion to solve for time:
h = ut + (1/2)gt^2
Where:
h is the height
u is the initial velocity (0 m/s as the block falls from rest)
g is the acceleration due to gravity (-9.8 m/s^2)
t is the time
Rearranging the equation, we get:
t = sqrt(2h / g)
Plugging in the values, we have:
t = sqrt(2 * 15.7 / 9.8)
≈ sqrt(3.22)
≈ 1.80 s
So the maximum time the man has to get out of the way is approximately 1.80 seconds.