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 February 8th, 2012
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.