Asked by Anthony
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
Answered by
Damon
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
Answered by
Damon
last two lines
d = 4.9 t^2
= 2.33 meters
d = 4.9 t^2
= 2.33 meters
Answered by
Anthony
TY very much Damon
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.