Question
a brick is dropped form the roof of a building. On the way down it passes a 2m high window and is observed to pass from the top to the bottom of the window in 0.25s. (a) How fast was it moving when it passed the top of the window? (b)How far below the roof of the building is the top of the window?
Answers
average speed passing window
= -2 meters/.25 s = -8 ms
constant acceleration = -9.81 m/s
so while crossing the window
v = Vi -.981t
change in v while crossing window = -9.81*.25 = -2.4525 m/s
half that = -1.23 m/s
so v at window top
= -8+1.23 = -6.77 m/s
so how far did it fall from the roof to the window top?
v = 0 -9.81 t
-6.77 = -9.81 t
t = .69 seconds
d = .49 t^2
=.233 meters
= -2 meters/.25 s = -8 ms
constant acceleration = -9.81 m/s
so while crossing the window
v = Vi -.981t
change in v while crossing window = -9.81*.25 = -2.4525 m/s
half that = -1.23 m/s
so v at window top
= -8+1.23 = -6.77 m/s
so how far did it fall from the roof to the window top?
v = 0 -9.81 t
-6.77 = -9.81 t
t = .69 seconds
d = .49 t^2
=.233 meters
last two lines
d = 4.9 t^2
= 2.33 meters
d = 4.9 t^2
= 2.33 meters
TY very much Damon