Find initial height when velocity is given

[KendraSan]

Homework Statement

A flowerpot falls from the ledge of an apartment building. It takes .2 s for the pot to pass the 4 m window below. How far above the top of the window is the ledge from which the pot fell? (Neglect air resistance)

Homework Equations

y(t)=y0 + (voy)(t) - 1/2g(t^2)
vy(t) = voy - gt

The Attempt at a Solution

4m/.2s =20m/s
so vy(.2)=voy-(9.81*.2)
20+1.962= voy
22= initial velocity

y(t)=y0 + 22(.2)-(9.81/2)(.2^2)
0 = y0 +4.4 - .2



Yeah, I know I'm wrong I just can't figure out how to do this problem.

 

[Hootenanny]

Homework Statement

A flowerpot falls from the ledge of an apartment building. It takes .2 s for the pot to pass the 4 m window below. How far above the top of the window is the ledge from which the pot fell? (Neglect air resistance)

Homework Equations

y(t)=y0 + (voy)(t) - 1/2g(t^2)
vy(t) = voy - gt

The Attempt at a Solution

4m/.2s =20m/s
so vy(.2)=voy-(9.81*.2)
20+1.962= voy
22= initial velocity

y(t)=y0 + 22(.2)-(9.81/2)(.2^2)
0 = y0 +4.4 - .2
Yeah, I know I'm wrong I just can't figure out how to do this problem.

You need to think about the pot's initial velocity again. What happens when you drop something? How fast it is going the moment you let go of it?