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?

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  

muscle endurance

explanation:

i can do 50 in a row ( but i get so tired after that)

it would be c. predictability