A flowerpot falls off a windowsill and passes the window of the story below. Ign

Charles Cisneros

Charles Cisneros

Answered question

2021-11-23

A flowerpot falls off a windowsill and passes the window of the story below. Ignore air resistance. It takes the pot 0.380 s to pass from the top to the bottom of this window, which is 1.90 m high. How far is the top of the window below the windowsill from which the flowerpot fell?

Answer & Explanation

Opeance1951

Opeance1951

Beginner2021-11-24Added 26 answers

We have the time the pot takes to fall a distance yy0=1.9 mm and g=9.8, from this we can calculated the velocity of the pot when it just reached the top of the windows in question. Take the positive y direction to be downward
yy0=v0t+12>2
v0=yy012>2t
v0=(1.9)(4.9×0.382)0.38=3.138 ms
The flowerpot falls from rest and we have the velocity it gains until it reaches the top of the window below, from this infirmation we can get the distance between the sill and the top of the window below.
v2=v02+2g(yy0)
yy0=v2v02g=3.138202×9.8=0.5 m
Result: The top of the windows is 0.5 m below the sill of the upper window

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