Free Fall Homework: Distance Between Window & Building

[veronicak5678]

Homework Statement

A brick falls from the top of a building and drops past a 2.0 m tall window in 0.20 s. Determine the distance between the top of this window and the top of the building.

Homework Equations

v^2 = (v initial)^2 + 2ax

The Attempt at a Solution

10 ^2 = 2(9.8)x

x= 5.102 m


Not at all sure if I used this formula correctly. Seems like the distance is too far.

 

[LowlyPion]

Homework Statement

A brick falls from the top of a building and drops past a 2.0 m tall window in 0.20 s. Determine the distance between the top of this window and the top of the building.

Homework Equations

v^2 = (v initial)^2 + 2ax

The Attempt at a Solution

10 ^2 = 2(9.8)x

x= 5.102 mNot at all sure if I used this formula correctly. Seems like the distance is too far.

Unfortunately, you can't just use the average Velocity like that.

$$v^2 = v_0^2 + 2 a \Delta x$$

Because in this equation the

$$v^2 - v_0^2$$

is not the square of the average velocity. It is the difference in the squares of the velocities.

Perhaps you should start from the

$$v = v_0 + a t$$

$$v_{bottom} = v_{top} + (9.8)*(.2)$$

 

[Kurdt]

From the question you were given a time, so it is safe to assume you need a kinematic equation with time in it. You also need one with distance and you know acceleration is that of gravity. It is always useful to review what quantities you are given to see what is best to use.

 

[veronicak5678]

So using v = v initial + at, I get 19.6 m/s. Do you mean that is the velocity at the bottom of the window?

 

[LowlyPion]

So using v = v initial + at, I get 19.6 m/s. Do you mean that is the velocity at the bottom of the window?

Yes I did mean V at the bottom of the window.

But what I was suggesting was that you can exploit the relationship between Vtop and Vbottom to solve the other equations.

By substituting

$$v_{bottom} = v_{top} + a* t$$

into

$$v_{bottom}^2 - v_{top}^2 = 2 a \Delta x$$

That way you can solve for an actual Vtop and then the rest is easy.

 

[veronicak5678]

OK. I used (18.57)^2 = 2(9.8) x to get x= 17.593 m. So the distance from the top of the building to the window top is 17.59 m.

 

[alphysicist]

Hi veronicak5678,

OK. I used (18.57)^2 = 2(9.8) x to get x= 17.593 m. So the distance from the top of the building to the window top is 17.59 m.

How did you get the speed of 18.57 m/s? That does not look right to me.

 

[LowlyPion]

OK. I used (18.57)^2 = 2(9.8) x to get x= 17.593 m. So the distance from the top of the building to the window top is 17.59 m.

Sorry, I don't get that at all.

I used Vbottom = Vtop +(9.8)(.2) = Vtop + 1.96

Substituting then into

$$v_{bottom}^2 - v_{top}^2 = 2 a \Delta x$$

$$(v_{top} + 1.96)^2 - v_{top}^2 = 2 (9.8) (2)$$

The v² cancels and you get a value for Vtop. Then you can use the same formula again but this time you solve for x which is your answer.

 

[veronicak5678]

I see what you mean. I made a silly math mistake in a hurry. Thanks for helping!