Physics
posted by Zalak on .
The Flower Pot: A flower pot falls off a windowsill and falls past the window below. It takes .3 s to travel from the top of the window to the bottom of the window which is 2.2 m tall. From what height above the top of the window did the flower pot fall?

Write equations for the times (t1 and t2) that it takes to pass the top and the bottom of the window, in terms of the unknown distance h.
t1 = sqrt(2h/g)
t2 = sqrt[2(h+2.2)/g]
Then, using those equations, write an equation for the time it takes to go from top to bottom of the window (which you know = 0.3s), in terms of the unknown h.
t2  t1 = 0.3
= sqrt[2(h+2.2)/g]  sqrt(2h/g)
Solve for h. 
0

h=ut +1/2at^2.
2.2=u*.3+1/2*10*(.3)^2
therefore,u=2.20.45/.3=5.8.
Taking 'u'=0 and v=5.8m/s.using the equation v^2=u^2+2as,we can caculate 's'=v^2/2a=1.682m. 
all wrong answers

In order to find the distance between the the bottom of the window above which is where the flowerpot was dropped(window 1) to the top of the window below (window 2) we need to find the velocity of the flowerpot as it passes the top of the window. Later this will serve as our "Vfinal." We know three things which is all you need to solve kinematic equations. 1) the distance the flowerpot travels = 2.2. 2) The time it took for the pot to travel 2.2 m= 0.3 seconds. 3) since this is a free fall problem acceleration and the flower pot is falling in the Ydirection ay = +9.8
we use this to solve for Voy at the top of the window 2
y=y0 + V0yt * 1/2ayt^2
Isolate Voy:
V0y= Dy/t  1/2ayt
V0y=2.2/0.3  1/2*9.8*o.3 = 5.86 m/s
now that you have your Vinitial, V0: solve for the distance between window 1 and window 2. We know three things: 1) since the flowerpot appears to have started from rest the initial velocity is zero and the initial velocity you just found becomes your finally velocity since you are finding the distance from the bottom of window one to the top of window two. 2) free fall problem = acceleration = 9.8, 3) final velocity = 5.86 m/s
now use kinematic equation to solve for distance deltay or "Dy"
Vf^2 = V0^2 + 2*ay*Dy
Dy = (Vf^2  V0^2)/ 2*ay
Dy = (5.86^2  0)/ 2 * 9.8 = 1.75 m
Vf is final velocity
V0 is initial velocity
Dy or Delta Y refers to the change in y or displacement. Change is y is similar to change in X. they are both displacement vectors but along a different axis or direction which is why it is very important to know the difference when your solving a physics problem. is the displacement occuring in the X or Y direction?