### An apartment-dweller is looking out of the window of her living room. The glass?

An apartment-dweller is looking out of the window of her living room. The glass

area of the window measures 2.00 m from top to bottom. A freely-falling computer monitor

suddenly drops past her window, taking exactly 0.200 s to pass the window. How far above

the top of the window was the monitor released (from rest) by the frustrated computer

user on a higher floor?

Highway

on November 30th, -0001

let the distance above the BOTTOM of the window be x, for now. (later, we’ll call it the midpoint of the window)

x = v(i)*t + 1/2at^2

where a = acceleration due to gravity = 10 m/sec^2

v(i) = initial velocity = 0 m/sec (it was dropped, not thrown)

the average velocity (v) in the latest interval of interest,

is 2.00 meter/ 0.200 sec = 10 meters/sec.

v = 1/2 a t

10 m/sec = 1/2 * 10 m/sec^2 * t

solve for t

t = 10 m/sec / 1/2 / 10 m/sec^2

t = 2 sec

the average velocity was when the monitor was midway

past the window, 1 meter from its top, 1 meter from its bottom.

and we know it took 2 seconds to get there.

we can solve for x where x is the distance to that midway point in the mirror, where we were at 10m/sec velocity

x = 1/2 a t^2

x = 1/2 (10m/sec^2)((2 sec) ^ 2))

x = 5*4 m

x = 20 meters, above the middle of the window

window is 2.00 m top to bottom

so the monitor was dropped 20m-(1/2*2m) =

19 meters above the top of the window.

This may not be the most direct method to get your answer, and I may have made some calculation errors.

So you’ll need to DOUBLE CHECK all the formulae, calculations and assumptions.

My disclaimer: it’s been 35 years since I’ve taken any math or physics course. Anyhow, this is my story, and I’m sticking to it.

I hope you find some of this helpful.