A cement block accidentally falls from rest from the ledge of a 52.1-m-high building. When the block is 19.8 m above the ground, a man, 2.00 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?
A locomotive is accelerating at 3.10 m/s2. It passes through a 25.0-m-wide crossing in a time of 3.03 s. After the locomotive leaves the crossing, how much time is required until its speed reaches 25.7 m/s?
To solve this problem, we need to use the equations of motion and kinematics.
First, let's determine the time it takes for the block to fall from a height of 52.1 m to a height of 19.8 m. We can use the equation:
Δy = ½ * g * t^2
Where:
Δy is the change in height (52.1 m - 19.8 m = 32.3 m)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time we're trying to find
Plugging in the values, we get:
32.3 = ½ * 9.8 * t^2
Rearranging the equation, we have:
t^2 = (32.3 * 2) / 9.8
t^2 = 6.58
t ≈ √6.58
t ≈ 2.56 seconds
So, it takes approximately 2.56 seconds for the block to fall from a height of 52.1 m to a height of 19.8 m.
Next, we need to determine the time it takes for the block to fall from a height of 19.8 m to the ground. We can use the same equation:
Δy = ½ * g * t^2
Where:
Δy is the change in height (19.8 m)
g is the acceleration due to gravity (9.8 m/s^2)
t is the time we're trying to find
Plugging in the values, we get:
19.8 = ½ * 9.8 * t^2
Rearranging the equation, we have:
t^2 = (19.8 * 2) / 9.8
t^2 = 4
t = √4
t = 2 seconds
Therefore, it takes 2 seconds for the block to fall from a height of 19.8 m to the ground.
Since the man noticed the block when it was 19.8 m above the ground, the maximum time he has to get out of the way is 2 seconds.