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Ber [7]
3 years ago
7

A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes the pot 0.380 ss to

pass from the top to the bottom of this window, which is 2.00 mm high.
How far is the top of the window below the windowsill from which the flowerpot fell?
Physics
1 answer:
kirill [66]3 years ago
7 0

Answer:

The top of the window is 0.5 m below the sill of the upper window  

Explanation:

We have the time the pot takes to fall a distance y- yo = 2.0 m and g = 9.8, from this we can calculate the velocity of the pot when it just reached the top of the window in question. Take the positive y direction to be downward  

y-yo=vo*t+I/2gt^2

  vo=(y-yo-I/2gt^2)/t

  vo=3.138 m/s

The flowerpot falls from rest and we have the velocity it gains until it reaches the top of the window below, from this information we can get the distance between the sill and the top of the window below.  

v^2=vo^2+2g(y-yo)

y-yo=(v^2-vo^2)/2g

      = 0.5 m

The top of the window is 0.5 m below the sill of the upper window  

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f': frequency when the train is getTing closer.

Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:

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