Help Please: a falling pot takes 0.2 s to fall past a window

I know this is a kinematics question but I'm having trouble starting it; Any help would be appreciated. (acceleration is assumed to be 9.8m/s (squared))

Question: A Falling flower pot takes 0.20 s to fall past a window that is 1.9 m tall. From what height above the top of the window was the flower pot dropped?

t= 0.20 s

d=1.9 m

a= 9.8 m/s (squared)

v1= 0m/s

Re: Help Please: a falling pot takes 0.2 s to fall past a wi

Maryanne said:

I know this is a kinematics question but I'm having trouble starting it; Any help would be appreciated. (acceleration is assumed to be -9.8m/s (squared))

Question: A Falling flower pot takes 0.20 s to fall past a window that is 1.9 m tall. From what height above the top of the window was the flower pot dropped?

Click to expand...

Sorry for the misdirection. I used 32 fps^2 for gravity instead of the metric 9.8mps^2

From h = Vot + gt^2, 1.9 = Vo(.2) + 4.9(.2)^2 from which we get Vo at the top of the window = 8.52m/s
om Vf = Vo + gt, 8.52 = 0 + 9.8t, making t from the drop point to the top of the window = .829 sec..

The height of the drop point above the window top derives from h = Vot + 16t^2 = o + 4.9(.869)^2 = 3.7m

Thanks Tchrwill, but the answer at the back of the book said 3.7 m......

\(\displaystyle \L \Delta x = v_0t - \frac{1}{2}gt^2\)

\(\displaystyle \L -1.9 = v_0(0.2) - 4.9(0.2)^2\)

\(\displaystyle \L v_0 = -8.52 m/s\) ... the pot's velocity at the top of the window

now, from where it was dropped to the top of the window ...

\(\displaystyle \L v_f^2 = v_0^2 - 2g(\Delta x)\)

\(\displaystyle \L \Delta x = \frac{v_f^2 - v_0^2}{-2g}\)

\(\displaystyle \L \Delta x = \frac{(-8.52)^2 - 0}{-19.6}\)

\(\displaystyle \L \Delta x = -3.7 m\)

so ... the distance from the point where the pot was dropped to the top of the window is 3.7 m.