An object is dropped from rest at a height of 112 m. Find the distance it falls during its final second in the air.
m
To find the distance the object falls during its final second in the air, we need to first find the time it takes for the object to fall from the starting height of 112 m to the height of 111 m, and then subtract this time from 1 second.
We can use the equation of motion for an object in free fall:
d = ut + (1/2)gt^2
Where:
d = distance
u = initial velocity (which is 0 since the object is dropped from rest)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Rearranging the equation to solve for time (t), we get:
t = √(2d/g)
The object falls from a height of 112 m to 111 m, so the distance (d) is 1 m. Plugging in the values, we get:
t = √(2 * 1 / 9.8)
t ≈ √0.204
t ≈ 0.452 seconds
Now, to find the distance the object falls during its final second, we subtract the time it took to fall 1 meter from 1 second:
Final time = 1 second - 0.452 seconds
Final time ≈ 0.548 seconds
Using the equation of motion, we can find the distance (d) during the final second:
d = ut + (1/2)gt^2
Since the initial velocity (u) is 0, the equation simplifies to:
d = (1/2)gt^2
Plugging in the values, we get:
d = (1/2) * 9.8 * (0.548)^2
d ≈ 0.15 meters
Therefore, the object falls approximately 0.15 meters during its final second in the air.