How do I solve this arduous physics problem? ?

Question

A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.24 s for the tile to pass her window, whose height is 2.5 m. How far above the top of this window is the roof?

_____m

Please help me! My friends and I have been working on this for hours and have no idea and it's due tomorrow!! :OOO

Answer 1

Let x be the distance from the roof to the top of the window, and x+2.5 be the distance to the bottom.  Let t be the time for the tile to reach the top of the window, and t+0.24 be the time to reach the bottom.  We then have:

x = 0.5(9.8) t^2

x+2.5 = 0.5(9.8)(t+0.24)^2

These can be solved by substitution of x from the first equation into the second, and dealing with the resulting algebra.  You first get the time, and from that get the distance.

Answer 2

You don't know the starting height, but you know the change in height from the top of the window, which is the reference point the answer wants. So make that equal to zero:

x‚ = 0

That makes the bottom of the window be negative (below 0):

x = -2.5 m

You have the time it takes to travel that distance:

t = 0.24 s

You know the acceleration, since that is gravity (negative, downward):

a = -9.8 m/s²

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The "falls from rest" part tells you the velocity at the top, but that isn't where our reference point is. So we are going to have to make another formula in order to use:

v = 0 m/s

That makes this a multi-part problem.

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You can't solve for at the top yet without knowing more about the velocity at the bottom. So you want to find velocity at the bottom:

v, = ?

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Find the equation with all of these things in it. It would take one of these idential forms:

Δx = v,·t + ½·a·t²

x - x, = v,·t + ½·a·t²

x = x, + v,·t + ½·a·t²

I'll use the last one (probably the most common). (Note though how the first one has "delta x", a change in x. That's what we are looking at here, a change in height from the top of the window to the bottom of the window.)

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Subtitute in for everything but v, which you don't know:

x = x, + v,·t + ½·a·t²

(-2.5 m) = (0) + v,·(0.24 s) + ½·(-9.8 m/s²)·(0.24 s)²

(-2.5 m) = (0.24 s)·v, + (-0.28224 m)

(-2.5 m) - (-0.28224 m) = (0.24 s)·v,

(-2.21776 m) = (0.24 s)·v,

(-2.21776 m) / (0.24 s) = v,

-9.24 m/s ≈ v,

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Now that you know v, you can solve for the position at the top. You want to find the time needed to fall that distance using:

v = v, + a·t

Remember that the velocity at the top was 0 m/s and you just found the velocity afterwards.

( -9.2 m/s ) = ( 0 m/s ) + ( - 9.8 m/s )·t

( -9.2 m/s ) = ( - 9.8 m/s )·t

( -9.2 m/s ) / ( - 9.8 m/s ) = t

0.94 s ≈ t

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Now just use the first formula again to find the position at the top.

x = x, + v,·t + ½·a·t²

( 0 m ) = x, + ( 0 m/s )·(0.94 s) + ½·( -9.8 m/s² )·(0.94 s)²

( 0 m ) = x, + ( -4.32964 m )

4.3 m ≈ x,

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Answer:

4.3 m above the top of the window.

Answer 3

acceleration = a = g = 9.8 m/s/s  

v2 - v1 = at

We know V1 =  0

Simple?