a flowerpot falls from a window ledge far above the street. Some distance down, phil fizyks observes that the pot takes 0.25 seconds to pass her window, which has a height of 2.0m. How far above the top of the window did the pot fall from?
find the velocity (window height/time) at the window.
vf^2=2g d solve for d
facts
To determine the distance the flowerpot fell from above the top of the window ledge, we can use the equations of motion under constant acceleration. In this case, we can assume that the only force acting on the flowerpot is gravity, which provides a constant acceleration of 9.8 m/s² downward.
First, we need to find the time it takes for the flowerpot to fall from the top of the window ledge to the window. We are given that it takes 0.25 seconds for the pot to pass the window.
Using the equation of motion: h = (1/2) * g * t², where h is the distance fallen, g is the acceleration due to gravity, and t is the time, we rearrange the equation to solve for t:
t = √(2h/g)
Substituting the given values: t = √(2 * 2.0 m / 9.8 m/s²)
Calculating: t = 0.645 seconds (rounded to three decimal places)
Now we can find the distance the pot fell from above the window ledge to the window by using the equation of motion: h = (1/2) * g * t².
Substituting the calculated value of t: h = (1/2) * 9.8 m/s² * (0.645 seconds)²
Calculating: h ≈ 2.00 meters (rounded to two decimal places)
Hence, the flowerpot fell from approximately 2.00 meters above the top of the window ledge.