Solving for Height of Falling Flowerpot

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SUMMARY

The problem involves calculating the height from which a flowerpot falls, given that it passes a two-meter tall window in ¼ of a second. The initial velocity (v0) is zero, and the velocity while passing the window is calculated using the formula v = d/t, resulting in 8 m/s. The correct approach to find the height involves using the equation h = 1/2 * g * t^2, leading to a final height of approximately 3.27 meters when substituting the appropriate values for gravity (g = 9.8 m/s²) and time (t = 0.25 s).

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Homework Statement



A flowerpot falls past two meter tall window in ¼ of a second. From what height did the flowerpot fall?

Homework Equations



To calculate the velocity of the pot while this is passing the window: v = d/t
To calculate the height: v^2= v0^2 = 2gh → h= (v^2 - v0^2) / 2g

The Attempt at a Solution



v = d/t = 2m / (1/4s) = 8 m/s

Since the initial velocity is 0, do we plug the 8m/s in place of the v squared, so the equation would be h= (8 m/s)^2 - 0 / 2 * 9.8 m/s^2 and the final answer should be 3.27m, right??
 
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While crossing the window, the flower pot is not moving with constant velocity. So your calculation of velocity is wrong.
Initial velocity of the flower pot vo = 0.
If the top edge of the window is at a height h, then
h = 1/2*g*t^2.
The bottom of the window is at a height h' =...?
Now h' = 1/2*g*(...)^2 ?
 

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