# Calculating Velocity Components and Slope for a Falling Rock in a Tunnel

• dwinkley
In summary: You didn't answer my question. Does the vertical component of the velocity increase or decrease as time goes by? You should also consider answer the question by @jbriggs444 because it is related.At first, the rock would be moving downwards at a very high speed, but as it gets closer to the ceiling, the speed decreases until it nearly stops at the closest point.
dwinkley
Homework Statement
(in this problem, i will be referring to this illustration i made: https://www.geogebra.org/calculator/ykyuetcp). We have the two red lines, which are the ceiling, and the floor of a tunnel. these lines are parallel to each other, and their difference in degrees from the horizontal axis is 30°. We have a random P point on the floor of the tunnel, and in the P point we have a rock(its size is the size of a point). The starting velocity of the rock is completely horizontal, there is no vertical starting velocity. The distance between the floor and the ceiling is 3 units. We want the rock to not hit the ceiling, but after lifting off the ground, we want it to hit the floor again as late as possible(we want to maximize the distance it travels, without it hitting the ceiling). What should its starting velocity be, and how far will it travel maximally?
Relevant Equations
vy=0
vx=?
all i could accomplish was calculating the distance between P and the ceiling in a horizontal line(6)

According to our rules, to receive help, you need to show some credible effort towards answering the question(s). How about telling us what you do know and how you would approach this problem. Vy=0 and vx=? don't say much. Also, it would be nice if you could post the figure you made here. Use the "Attach files" link on the lower left.

Grelbr42
kuruman said:
According to our rules, to receive help, you need to show some credible effort towards answering the question(s). How about telling us what you do know and how you would approach this problem. Vy=0 and vx=? don't say much. Also, it would be nice if you could post the figure you made here. Use the "Attach files" link on the lower left.

the only actually useful thing i managed to figure out was that the distance between p and the ceiling is 6 units horizontally. since this is a problem way above my current physics knowledge, i cant do much to solve it, its just that im very interested in the solution of it, since i find the problem itself interesting as well. i saw it in a physics competitions final round if you are interested in where i found it. i attached the figure i made with geogebra here. all non-colored lines are just helper lines, and are not really relevant to the problem, but just helped me illustrate things accurately.

#### Attachments

• Physics Problem.png
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The key to this problem is translating into math the condition that the rock not hit the ceiling and travel as far as possible. "Not hit the ceiling" means "get as close to the ceiling as possible" which can be taken as zero distance. Of course the rock is still moving at that closest point. You know that the horizontal component of the velocity is equal to the initial speed. What about the vertical component? What should it be at the point of closest approach?

Last edited:
i think it should be zero, since if it would be any larger, we would hit the ceiling shortly after

dwinkley said:
i think it should be zero, since if it would be any larger, we would hit the ceiling shortly after
If the rock is flying past, almost brushing against the ceiling, what direction must it be travelling?

dwinkley said:
i think it should be zero, since if it would be any larger, we would hit the ceiling shortly after
Can the vertical component get any larger? It starts at zero and then what happens? Does it increase or increase as time goes by?

the vertical component starts at zero, and ay is equal to g. if we round g to 9.8, then the vertical acceleration is 9.8 as well

You didn't answer my question. Does the vertical component of the velocity increase or decrease as time goes by? You should also consider answer the question by @jbriggs444 because it is related.

i said ay will be equal to g, meaning the y velocity will increase by g every second.

dwinkley said:
i said ay will be equal to g, meaning the y velocity will increase by g every second.
So you are saying that at the beginning of the second second, the rock is moving upward at 9.8 meters per second?

no, it will move downwards every second by that amount, but the amount with which it will move down is constantly increasing

dwinkley said:
no, it will move downwards every second by that amount, but the amount with which it will move down is constantly increasing
Recall that @kuruman had asked about the vertical component of velocity at the moment of closest approach to the ceiling.

You had suggested that that velocity was zero. Now you are suggesting that it is ##9.8 t## meters per second downward. But if we do not know the time of closest approach, that formulation does not help much.

Still, we are getting somewhere. You seem to agree that the rock is moving downward at some angle when it approaches the ceiling most closely.

What is that downward angle?

im sorry, when i said 0, i was just confusing vertical with horizontal(im not from an english speaking country)
my guess for the angle with which its travelling would be completely parallel to the ceiling, meaning a difference of 30 degrees from horizontal

dwinkley said:
im sorry, when i said 0, i was just confusing vertical with horizontal(im not from an english speaking country)
my guess for the angle with which its travelling would be completely parallel to the ceiling, meaning a difference of 30 degrees from horizontal
Good. I am not sure exactly how @kuruman was intending to proceed from here but...

Given this angle, if the horizontal velocity is denoted as ##v_x##, then what would ##v_y## have to be?

i believe its vx/sqrt(3). my calculations go as follows: we have vx, and vy(which are perpendicular to each other), and coming from the point, we have a v vector as well(the velocity as 1 variable), which closes a 30 degree angle with vx. if we connect the end of v to the end of vx, we get a semi-regular triangle(idk how you call it in english its basically a regular(equilateral) triangle cut in half, and from there, if we mirror v to vx, we can see that vx = vy*sqrt(3), and vy = vx/sqrt(3). hope im correct lol

While the distance traveled back to the bottom can be computed from the OP (distance in 'units'), we don't know what those distance units are. Meters? AU? Angstroms? The velocity question very much depends on what the 'unit' of distance is. Either that, or the gravity needs to be expressed in terms of distance units per time-units²

i just checked the original test that this problem was from, and it said on top that all things are measured in meters, and then proceeded to say "unit" in every problem. pretty dumb if you ask me, and seems completely unnecessary. they did the same thing with declaring time units at the top, volume units, etc

dwinkley said:
i believe its vx/sqrt(3).
Yes. Which is also equal to ##v_x \tan 30^{\circ}##

so where do we go from here?

dwinkley said:
so where do we go from here?
There are at least two ways to proceed. Probably several others. I have not figured out where @kuruman is heading with his approach. One way he might be going is this...

We have the initial vertical velocity. Zero.

We have the initial horizontal velocity. ##v_x## as a named unknown.

We have the final vertical velocity. ##\frac{v_x}{\sqrt{3}}##

That means that that we can compute the average horizontal velocity (trivial) and the average vertical velocity for the trajectory from launch to the point of closest approach with the ceiling for an optimally falling rock.

Importantly, we can compute the ratio of those average velocities. So we know the angle of the average velocity. So we can project a line from the launch point to the ceiling and figure out where the rock comes closest to the ceiling.

With that information in hand, we will have drop distance from launch to that point. That, in turn, gives us elapsed time.

With that information in hand, we will be able to compute launch velocity.

Edit: I see that @kuruman is going for a more algebraic approach while I have gone for a more physical approach.

MatinSAR
dwinkley said:
so where do we go from here?
What is the slope of the line for the ceiling?
Can you find an expression for the time at which the velocity components are such that ##\dfrac{v_y}{v_0}## is equal to the slope?
If so, you can use the kinematic equations to find the coordinates ##\{x_p,y_p\}## of the point of closest approach in terms ov ##v_0##.

Then you write a straight line equation for the ceiling, in the form ##y=ax+b##, substitute the values for ##x_p## and ##y_p## and you will get an equation that you can solve for ##v_0##.

MatinSAR and SammyS

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