- #1
dxlogan187
- 5
- 0
A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.19 s for the tile to pass her window, whose height is 1.4 m. How far above the top of this window is the roof?
My train of thought is:
You are given the initial Velocity (Vo=zero m/s), acceleration (-9.8 m/s^2), and the instantaneous velocity of when the tile passes the window (1.4/0.19 = 7.36 m/s)
So you use the instantaneous velocity and plug it into the equation
Vf=Vo+(a)(t) to find the amount of time it took to get to the window.
But if I do that, I get a negative time. I'm not sure where to go here because I can't proceed with a negative time.
After that, my idea was to take the time, and insert it into another equation and find the distance. Even if I neglect that I get a negative time, I still get the incorrect answer after plugging it into a different equation. So I think I have two problems I need to overcome.
Thanks for the help,
Logan
My train of thought is:
You are given the initial Velocity (Vo=zero m/s), acceleration (-9.8 m/s^2), and the instantaneous velocity of when the tile passes the window (1.4/0.19 = 7.36 m/s)
So you use the instantaneous velocity and plug it into the equation
Vf=Vo+(a)(t) to find the amount of time it took to get to the window.
But if I do that, I get a negative time. I'm not sure where to go here because I can't proceed with a negative time.
After that, my idea was to take the time, and insert it into another equation and find the distance. Even if I neglect that I get a negative time, I still get the incorrect answer after plugging it into a different equation. So I think I have two problems I need to overcome.
Thanks for the help,
Logan