How Far Does a Rock Travel Horizontally When Thrown from a Roof?

[RyanJF]

Homework Statement

"A rock is thrown horizontally at 25 m/s from a 15-m roof. How far does the rock travel away from the building before hitting the ground?"

In the X-component:

Xx = ?
Vox = 25 m/s
T = ?
Vx = 0

In the Y-component:

Xy = 15m
Ag = 9.8
T = ?
Vy = 0

Homework Equations

Kinematics equations

The Attempt at a Solution

First thing I did was picked an appropriate equation that would allow me to solve for Xx. The equation that I selected was:

Xx = 1/2 (Vox + Vx) T

In order to solve for Xx, it would first be necessary for me to acquire a value for T. I solved for T by doing:

T = sqrt(2x/a)

Plugged in with the correct numbers, this turned out to:

T = sqrt(2*15/9.8)

Therefore

T = 1.75 sec

I would assume that I'm using the vertical value for distance (15 m), given that that downwards motion, in conjunction with acceleration due to gravity, shall determine how far the projected rock will travel.

My final answer ended up being:

Xx = 21.875 m

Although I think the correct answer should be closer to around 43.74 m. This was just an example problem given to the class, but I'm not sure how I'm supposed to arrive at the presented conclusion. I'm quite confident that I correctly solved for T, and I thought I used the right equation for displacement, but there's clearly something that I did wrong.

Can somebody please explain this to me?

 

[collinsmark]

First thing I did was picked an appropriate equation that would allow me to solve for Xx. The equation that I selected was:

Xx = 1/2 (Vox + Vx) T

Sorry, but that's not the correct equation. As a matter of fact, it's not even one of your kinematics equations at all. Try again with something else.

[Edit: Okay, I think I see where this formula came from now. It was the notation that threw me. I think you mean:
X_x = (1/2)(V_ox + V_fx)t,
which is fine. But realize that in this particular problem, V_ox = V_fx, and neither of them is zero!]

In order to solve for Xx, it would first be necessary for me to acquire a value for T. I solved for T by doing:

T = sqrt(2x/a)

Plugged in with the correct numbers, this turned out to:

T = sqrt(2*15/9.8)

Your calculation for t is correct! However, I would have used a y in the equation (or X_y, to follow your notation) instead of x, but now I'm just being nit-picky. Whatever the case, you got the right value for time t.

My final answer ended up being:

Xx = 21.875 m

Although I think the correct answer should be closer to around 43.74 m.

Try it again with the correct formula for X_x.

 

[RyanJF]

Thanks, Collinsmark!

Normally I would found the alt codes to insert the proper subscripts, but I made this thread at something like twelve in the morning, so I was in somewhat of a rushing, hoping both to have an answer in the morning and to get to sleep.

I understand now how to work the equation (I typically don't have much of a problem with the math, just some of the integration), but how come the initial velocity and final velocity equal one another? I assumed that the final velocity would have been equivalent to zero, since the rock has no velocity at the end of the "story", as it has hit the ground.

Even if that were not the case, wouldn't the rock speed up because gravity would accelerate it? Or, because the problem deals only with the horizontal component, does Vx remain the same?

Hm. If the assumption I made above remains the same, would that mean that, in a similar situation, Vy would equal zero, because it's dealing with the vertical component, and the vertical component would be what's falling (thus equaling zero once hitting the ground)?

Apologies for the notation. It's what my physics book uses, I don't much care for it, myself.

 

[collinsmark]

I understand now how to work the equation (I typically don't have much of a problem with the math, just some of the integration), but how come the initial velocity and final velocity equal one another?

In the particular part of the problem we are talking about, there is no acceleration in the horizontal direction (x-component). There is acceleration in the vertical direction (y-component), but we were discussing the horizontal direction.

If you take the horizontal direction (x-component) by itself, there is no acceleration, no force. Newton's first law of motion states, "an object in motion tends to stay in motion unless acted upon by an outside force." Since there are no forces in the x-direction,...

I assumed that the final velocity would have been equivalent to zero, since the rock has no velocity at the end of the "story", as it has hit the ground.

Don't take it that far. You're looking for parameters immediately before it hits the ground.

Even if that were not the case, wouldn't the rock speed up because gravity would accelerate it?

In the y-direction, yes. But not the x-direction. The acceleration is zero in the x-direction.

Or, because the problem deals only with the horizontal component, does Vx remain the same?

There you go. Keep in mind the rock moves in the x-direction though. I.e. it has a velocity. But it doesn't accelerate in the x-direction.

Hm. If the assumption I made above remains the same, would that mean that, in a similar situation, Vy would equal zero, because it's dealing with the vertical component, and the vertical component would be what's falling (thus equaling zero once hitting the ground)?

Again, you're concerned with the parameters of the rock just before it hits the ground. Not after. Just before the rock hits the ground, the y-component of velocity is definitely non-zero.