How Fast Will the Object Be Falling When It Passes Ken's Window?

[mt05]

hey everyone! I think this question can probably be solved using a constant acceleration formula, but since we are doing the momentum and work units in class i figured i was suppose to use those formulas to solve my question. anyways here it is :

Paul lives on the sixth floor of an apartment complex. His window is 20.2 m above the ground. Paul notices a 7.25 kg object falling past his window at 8.50 m/s. If Ken's window is 5.00m above the ground level, how fast will that same object be falling as it passes by Ken's window?

KE = 1/2(7.25)(8.50)^2
KE = 261.91
KE = W, W = FD, FD = 1/2mv^2final - 1/2mv^2initial

261.91 = 1/2(7.25)(v^2) - 1/2(7.25)(8.50)^2
261.91 = 3.625v^2 - 261.91
523.82/3.625 = v^2
vfinal = 12 m/s.
i know that is wrong, because when i put the numbers into the formula vf^2 = vi^2 + 2ad i get 19.2 m/s which seems more reasonble. Is there anyway to figure this question out properly by using momentum/work ?
KE = W, W = FD, FD = 1/2mv^2final - 1/2mv^2initial ( this doesn't seem to make much sense either, but i thought i saw it on my formula sheet )
thanks for any help!

 

[Redbelly98]

Welcome to PF

KE = 1/2(7.25)(8.50)^2
KE = 261.91
KE = W, W = FD, FD = 1/2mv^2final - 1/2mv^2initial

261.91 = 1/2(7.25)(v^2) - 1/2(7.25)(8.50)^2
261.91 = 3.625v^2 - 261.91

Work W is not equal to the initial KE, so it's wrong to simply plug in the initial KE of 262 J for W.

Instead, calculate what W=FD is by figuring out the values of F and D.