Solve the Tough Question: Roof Tile Falling from a Building

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Homework Help Overview

The problem involves a roof tile falling from rest from the top of a building, with an observer noting the time it takes for the tile to pass a window. The context is kinematics, specifically dealing with uniform acceleration due to gravity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial conditions of the tile's motion, questioning the assumption that the initial velocity is zero when the tile passes the observer. There are various calculations proposed to determine the height of the roof above the window, with some participants suggesting different equations and methods to find the initial velocity and distance.

Discussion Status

The discussion is active, with multiple interpretations of the problem being explored. Some participants have provided hints and partial calculations, while others are questioning the assumptions made about the initial velocity of the tile. There is no explicit consensus on the approach to take, but several lines of reasoning are being examined.

Contextual Notes

Participants note the specific value of gravitational acceleration used in Canada (9.81 m/s²) and the time it takes for the tile to pass the window (0.20 seconds). There is an emphasis on the need to consider the tile's velocity as it passes the top of the window, which is a key point of contention.

nofear001
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Ive been working on this and its a little tough: Whats the answer?

A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.20s for the tile to pass her window whose size is 1.6m. How far above the top of this window is the roof?
(Hint: 2 calculations- Use uniform acceleration equations)

How am i suppose to do this?

a=9.81m/s^2 (im in Canada we use 9.81 not 9.8)
t= 0.20s
d= ?
Vi= 0m/s
Vf=?

Thxs
 
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Hint: the initial velocity (that is, the velocity as it passes the top edge of the window) is not zero. You need to solve for this velocity, as it will tell you how far from the roof the top of the window is.

- Warren
 
but the rooftile falls from rest rest= 0m/s
 
is the answer 2.51 meters?
 
The roof tile is NOT AT REST when the observer first observes it.

First, it falls from the roof to the top of the window, gaining some velocity. Next, it falls past the window, which the observer times at 0.2 seconds.

Its velocity, when the observer first sees it, is NOT zero. You need to solve for this velocity first. You can then use it to solve for the height of the roof above the window.

- Warren
 
Heres the caluclations

s = 1.6 t= .2 a = g = 9.81 v =?
s = vt +at^2
and v turns out to be 6.038
well this is the velocity when it reached the point -lets say- x which is when observer first sees it. because it fell off the roof and traveled let's say distance d already...so...
d = (v^.5)/2g
= 1.86 m

i think that's pretty much it.
 
s = (1/2)gt^2

and

s+1.6 =(1/2)g(t+0.20)^2

think! and eliminate t.

M.P.
 
Not going to give an answer here but the working I would use seems to differ slightly from others, and arrives at a slightly different expression.

we know that distance = 1/2(inital velocity + final velocity) x time or...
s =1/2(u+v)t
we also know that velocity (v) = u + at
using substitution you can end up with an equation that involves only one variable...that is 'u'
 

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