A student sees a flowerpot pass his 2m high window on the fourth floor in 0.093 seconds. The distance between floors is 4.0m. From which floor did the pot fall?
velocity of pot: 2/.093
Vf^2=2gh
solve for h, given Vf=2/.093
Then subtract 2 to get the height above the center of your window.
To determine from which floor the flowerpot fell, we can use the equation of motion for freely falling objects, which is:
h = 0.5 * g * t^2
Where:
- h is the vertical distance traveled (in meters)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken to fall (in seconds)
In this case, the student observed the flowerpot passing his 2m high window on the fourth floor in 0.093 seconds. The distance between floors is 4.0m.
Let's calculate the height from which the flowerpot fell:
1. First, find the total distance traveled by the flowerpot from the top of the fourth floor to the window height:
distance = 4th floor height + window height = 4.0m + 2.0m = 6.0m
2. Next, use the equation of motion to find the time it takes for the flowerpot to fall the whole distance between the floors:
h = 0.5 * g * t^2
4.0m = 0.5 * 9.8 m/s^2 * t^2
Solving for t, we get:
t^2 = (4.0m * 2) / (9.8 m/s^2)
t^2 = 0.8163
t ≈ √0.8163
t ≈ 0.904s
3. Finally, subtract the observed time (0.093s) from the time it takes for the flowerpot to fall between the floors (0.904s) to determine the time it took to fall from the top floor to the 4th floor:
Time taken to fall between floors = 0.904s - 0.093s = 0.811s
Now, we need to figure out how many floors the flowerpot passes in 0.811 seconds:
4. Calculate the number of floors crossed in the observed time:
Floors crossed = (distance between floors) / (time taken to fall between floors)
Floors crossed = 4.0m / 0.811s ≈ 4.93
Since the calculated number of floors crossed is approximately 4.93, we can conclude that the flowerpot fell from the 5th floor.