A cement block accidentally falls from rest from the ledge of a 52.0 -m-high building. When the block is 11.5 m above the ground, a man, 1.80 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 determine the maximum time the man has to get out of the way, we need to find the time it takes for the cement block to fall from a height of 11.5 m to the ground. We can use the concept of free fall and the equations of motion to solve this problem.
First, let's consider the time it takes for the block to fall from the top of the building to a height of 11.5 m. We can use the equation for the distance traveled during free fall:
y = (1/2) * g * t^2
Where:
- y is the vertical distance traveled (11.5 m)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken
Rearranging the equation to solve for time (t):
t = √(2y / g)
Substituting the given values:
t = √(2 * 11.5 / 9.8)
t = √(23 / 9.8)
t ≈ 1.52 seconds
Therefore, it takes approximately 1.52 seconds for the cement block to fall from a height of 52.0 m to a height of 11.5 m.
Now, let's find the time it takes for the block to fall from a height of 11.5 m to the ground. We can use the same equation:
t = √(2 * y / g)
Substituting the values:
t = √(2 * 11.5 / 9.8)
t = √(23 / 9.8)
t ≈ 1.52 seconds
Since the time taken for both portions of the fall is the same and the man notices the block when it is 11.5 m above the ground, the maximum time the man has to get out of the way is approximately 1.52 seconds.