How High Above the Window is the Roof?

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SUMMARY

The discussion centers on solving a physics problem involving a roof tile falling from rest and passing a window. The key parameters are a time of 0.26 seconds for the tile to pass a 1.6-meter window and an acceleration due to gravity of -9.8 m/s². The correct approach involves using the kinematic equation Δy = v0Δt + (1/2)at² to find the distance between the roof and the top of the window, resulting in a height of 0.3384 meters above the window. The negative displacement indicates the downward motion of the tile.

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Melody
Hello! I'm a new user. I'm currently a college student taking Physics. I have previous Calculus experience, so the mathematical area isn't what I'm having problems in.

There is a question that I was assigned that has got be running around in circles trying to answer. The question appears:

A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.26 seconds for the tile to pass her window, whose height is 1.6 meters. How far above the top of this window is the roof?

At the moment, I know that I have

time = 0.20 seconds
window height = 1.6 meters
gravity acc = 9.8 m/s^2

But I don't know where to go from here. I was trying to use the equation :

v = vknot + a(subscriptx) x delta t

But I don't know if that would even apply to this problem.

Thank you for any help, assistance, or method of starting this problem! :D
 
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You are told that the acceleration due to gravity is -9.8 m/s^2.

Since you have "calculus experience" you ought to know that acceleration is the derivative of velocity so you can find the velocity by integrating (in this simple case of constant acceleration, that is just multiplying): the velocity at time t (seconds) is v(t)= -9.8t+ v0. The problem says "falls from rest" so v0= 0. v(t)= -9.8t.
velocity is, of course, the derivative of the "position function" (in the case of something falling, we can take that to be "height above the ground") so height is the integral of the velocity function: h(t)= -4.9t^2+ h0.
We DON'T KNOW the initial height, that's what's asked so we just leave it as an unknown. Since that equation has three "unknowns" (time t, height h, and initial height h0), we will need to know some specific h and t in order to solve for h0.
Fortunately, we are told that "An observer inside the building notices that it takes 0.26 seconds for the tile to pass her window, whose height is 1.6 meters." So that we know that when t= 0.26, h= 1.6. Put those values into the equation to solve for h0.
Of course, the question doesn't ask for the height of the building! It asks "How far above the top of this window is the roof?"
You should be able to calculate that from the height of the roof and window above the ground.

By the way, you note: "I know that I have time = 0.20 seconds". First, of course, you've copied that wrong: the time at which the tile passes the window is .26, not .20! But also, it doesn't do any good to note "time= 0.26 seconds" without saying WHAT time that is: it is the time WHEN h= 1.6 m. That's why knowing "v = vknot + a(subscriptx) x delta t" doesn't help you. It's HEIGHT that is important, not v. Also, before you can use a formula, you have to know WHAT IT MEANS! It was not clear to me that you understood what "delta t" meant in this problem. And "vknot"? Did you mean v "naught"? It's better to just write v0 which, I suspect,was the way the formula was given in your book.
 


Hello! It sounds like you have a good understanding of the given information and are on the right track with using the equation v = v0 + aΔt. This equation, also known as the kinematic equation for displacement, can definitely be applied to this problem.

To start, it's important to identify what you are trying to solve for. In this case, you are trying to find the distance between the top of the window and the roof, which can be represented as Δy.

Next, plug in the known values into the equation. The initial velocity (v0) is 0 m/s since the tile is starting from rest. The acceleration (a) is -9.8 m/s^2, as it is acting in the opposite direction of the tile's motion. The time (Δt) is 0.26 seconds, as given in the problem.

Now, rearrange the equation to solve for Δy:

Δy = v0Δt + (1/2)at^2

Substituting in the known values, we get:

Δy = (0)(0.26) + (1/2)(-9.8)(0.26)^2

Solving for Δy, we get a value of -0.3384 meters. This means that the roof is 0.3384 meters above the top of the window.

One thing to note is that the negative sign indicates that the displacement is in the opposite direction of the acceleration, which makes sense since the tile is falling downwards.

I hope this helps you solve the problem! Remember to always identify what you are trying to solve for and use the appropriate equation to find the solution. Good luck with your studies!
 

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