A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes the pot 0.420 seconds to pass this window, which is 1.90m high. How far is the top of the window below the windowsill from which the flowerpot fell?
1) V= Vo + at
2) Y = Yo + VYot + (1/2)at^2
3) V^2 = Vo^2 + 2a(Y - Yo)
The Attempt at a Solution
I tried to solve the problem by setting the origin to be the ground, the height of the window as 1.90m, and the height of the initial position of the flowerpot at the windowsill as some unknown Yo (initial y). I believe the Vyo (initial velocity) should be zero. I then used the time .420 seconds to put into the position equation and tried to solve for Yo.
Y = Yo + VYot + (1/2)at^2
1.90m = Yo + 0(.420s) + (1/2)(-9.8 m/s^2)(.420s^2)
(Doing the algebra)
2.65m = Yo or the height of the windowsill
Next I subtracted 1.90m from 2.76m and found .86m to be the distance between the window and the above windowsill but this is incorrect. The answer is 0.31m by the way but I dont know how to get to that answer.
I believe there is something wrong with the way I am picturing and setting up the problem. Thank you very much to anyone who can help.