Location (x,y) when a Projectile Collides with a Slanted Wall

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NOTICE: THIS IS NOT AN ACTUAL HOMEWORK PROBLEM.

Homework Statement


A projectile is launched from a hill 2m high 8m away from the base of a slanted wall at a velocity v at an angle of 35 degrees from the horizontal. The equation of the slanted wall is y = 2.8x. The slanted wall is 20m high. Ignore air resistance/drag and use g = -9.8m/s2. a) Where will the object collide with the slanted wall? Give (x,y) coordinates. b) What is the minimum velocity of the object needed to clear the wall? c) Give a mathematical representation of this problem where the angle the object is launched at is unknown.


Homework Equations


s = s0 + v0t + 1/2gt2
http://en.wikipedia.org/wiki/Trajectory
http://en.wikipedia.org/wiki/Projectile_motion
http://en.wikipedia.org/wiki/Kinematics


The Attempt at a Solution


I tried to solve this by representing the wall and the trajectory of the projectile graphically, but I'm only in high school level Physics, so I was absolutely clueless as to what equation I would use to represent the trajectory of the projectile so that the x axis was x displacement and the y axis was y displacement. (I googled a bit) I figured that if I graphed both of those equations the intersection point would be where the projectile hits the wall (so long as I changed the equation of the wall so it's x intercept is 8). Anyways, that didn't work out, and I've been wracking my brain as to how to solve this for 4 hours, help please!

P.S.: This is not an actual homework or classwork problem, I came up with this problem myself because I was trying to explain something that I saw in real life. ;]

Here is a diagram of the problem that I made in paint:
[PLAIN]http://img220.imageshack.us/img220/4765/diagramr.png [Broken]

The red curved line is my attempt at a parabola representing the trajectory of the projectile in the circumstance where the object comes in contact with the wall 2m above the ground.

Thanks,
Mator
 
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Answers and Replies

  • #2
gneill
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You were on the right track with the equations of motion and the wall. You'll end up having to solve three simultaneous equations for the values of x,y, and t (you only seem to need x and y).

If you place the base of the slanted wall at the origin, then the gun is firing from:

x0 = -8m y0 = 2m

and the equation(s) of motion for the y and x values become:

y = y0 + v*sin(q)*t - (1/2)*g*t2

x = x0 + v*cos(q)*t

The wall front has the equation:

y = a*x

where a is the slope of the wall (2.8)

To solve, start by substituting the expression for the wall y value into the equation for the projectile's y value, then replace t in that equation with the t from a rearrangement of the projectile x value equation. You'll end up with an equation that only involves x as an unknown. This will end up being a quadratic equation. I might suggest making the substitution x = (x - x0) at this point, and solving the quadratic for x.

You might also want to check to see what the line-of-sight angle is between the gun and the top of the wall. How would this affect your minimum firing angle for reaching or exceeding it?
 
  • #3
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You were on the right track with the equations of motion and the wall. You'll end up having to solve three simultaneous equations for the values of x,y, and t (you only seem to need x and y).

If you place the base of the slanted wall at the origin, then the gun is firing from:

x0 = -8m y0 = 2m

and the equation(s) of motion for the y and x values become:

y = y0 + v*sin(q)*t - (1/2)*g*t2

x = x0 + v*cos(q)*t

The wall front has the equation:

y = a*x

where a is the slope of the wall (2.8)

To solve, start by substituting the expression for the wall y value into the equation for the projectile's y value, then replace t in that equation with the t from a rearrangement of the projectile x value equation. You'll end up with an equation that only involves x as an unknown. This will end up being a quadratic equation. I might suggest making the substitution x = (x - x0) at this point, and solving the quadratic for x.

You might also want to check to see what the line-of-sight angle is between the gun and the top of the wall. How would this affect your minimum firing angle for reaching or exceeding it?
Thank you so much for you fast and detailed response!

Ok, let's do this.

2.8x = 2 + v*sin(35)*t - (1/2)*g*t2
t = (x - 8)/(v*cos(35))

2.8x = 2 + ((v*sin(35))(x - x0))/(v*cos(35)) - (1/2)*g*((x - x0)/(v*cos(35)))2

2.8x = 2 + tan(35)(x - x0) - (1/2)*g*((x - x0)/(v*cos(35)))2

ok g2g to next class, will edit post later
 
  • #4
754
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Here is a diagram of the problem that I made in paint:
[PLAIN]http://img220.imageshack.us/img220/4765/diagramr.pngr[/QUOTE] [Broken]

I'm sure this diagram is correct, but as stated, there is nothing to say that the wall slants away from the projectile. (What if the projectile was shot from the right hand side of the wall?)
 
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  • #5
gneill
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20,816
2,792
I'm sure this diagram is correct, but as stated, there is nothing to say that the wall slants away from the projectile. (What if the projectile was shot from the right hand side of the wall?)
Then you would need to change the value of x0 and the equation for the wall facing the projectile (it would be the vertical backside) accordingly, and redo the equations.
 
  • #6
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I'm sure this diagram is correct, but as stated, there is nothing to say that the wall slants away from the projectile. (What if the projectile was shot from the right hand side of the wall?)
The equation for the wall would be y = -2.8x then, which it wasn't given as. Though I never considered that possibility because I made the problem myself. It was describing something I observed (as stated before). I might post a link to a video because it's hard to explain (if any is curious enough to watch a video).

OK. Well I managed to do the rest of all the work during lunch and in between classes. I'll try to make this as concise, yet understandtable, as I can:

x = x0 + vcos(q)t
y = y0 + vsin(q)t + 1/2gt2
y = mx

mx = y0 + vsin(q)t + 1/2gt2

(x-x0)/(vcos(q)) = t

mx = y0 + (vsin(q))*((x-x0))/(vcos(q)) + 1/2g((x-x0)/(vcos(q)))2

a = g/2
b = vsin(q)
c = y0

t = (x-x0)/(vcos(q)) = (-vsin(q) +/- sqrt(v2sin(q)sin(q) - 4(y0)(g/2)))/(g)

x = x0 + (-v2sin(q)cos(q) +/- vcos(q)*sqrt(v2sin(q)sin(q) - 4(y0)(g/2)))/(g)

y = x/m (just substitute x in)

Then it was just plug and chug for part b, for which I got 25.277m/s

:D

Thanks everyone, I think that's right, but checking it wouldn't hurt!

-Mator
 
  • #7
754
1
All I was saying is that there is nothing stating that the wall slopes away from the hill because there is no statement as to where the hill is in relation to the wall (only how far away it is).
 
  • #8
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All I was saying is that there is nothing stating that the wall slopes away from the hill because there is no statement as to where the hill is in relation to the wall (only how far away it is).
Oh well excuuuuuuse me for not making the problem ABSOLUTELY PERFECT. I added a friggin' diagram so did you really have to point out that tiny insignificant error in the problem? I wrote this myself, and it's the first problem I've ever written and it's nothing like what I'm learning in school. So if you have nothing to contribute just get out. Furthermore the diagram is correct because I made the problem. For christ sake how many times do I have to say that.

-Mator
 
  • #9
754
1
So you want help, but you don't want anybody to point out anything you did wrong? That's an interesting way to learn...



No need to take offense. I was simply pointing out that the problem could be interpreted 2 different ways, as stated.

The problem doesn't refer to the diagram; rather, as stated, your diagram is part of your solution.

It's a very good problem. I was just pointing out that there is a minor detail left out of the wording.



BTW, when dealing with physics and math - it's all about the details.

And if you aren't willing to take some constructive criticism, maybe you shouldn't post questions here.
 
  • #10
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So you want help, but you don't want anybody to point out anything you did wrong? That's an interesting way to learn...



No need to take offense. I was simply pointing out that the problem could be interpreted 2 different ways, as stated.

The problem doesn't refer to the diagram; rather, as stated, your diagram is part of your solution.

It's a very good problem. I was just pointing out that there is a minor detail left out of the wording.



BTW, when dealing with physics and math - it's all about the details.

And if you aren't willing to take some constructive criticism, maybe you shouldn't post questions here.
That was way too composed. Kudos to you, sir.
 
  • #11
754
1
By the way, there is another "tiny insignificant" problem...

part b) is impossible. At 35 degrees, the projectile cannot clear the wall. It can only reach a maximum vertical height of approximately 10.14 m up the wall. (Unless it is carried by an updraft!)
 
  • #12
11
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So you want help, but you don't want anybody to point out anything you did wrong? That's an interesting way to learn...



No need to take offense. I was simply pointing out that the problem could be interpreted 2 different ways, as stated.

The problem doesn't refer to the diagram; rather, as stated, your diagram is part of your solution.

It's a very good problem. I was just pointing out that there is a minor detail left out of the wording.



BTW, when dealing with physics and math - it's all about the details.

And if you aren't willing to take some constructive criticism, maybe you shouldn't post questions here.
No, I was fine with you pointing out that I made a minor oversight on a tiny insignificant (yes, insignificant because I provided a diagram and because I made the problem) detail that I missed. I wasn't fine with you repeating it over and over, bloating the minor oversight out of proportions. You are the one who didn't drop the subject after I recognized it.

I had the diagram in a P.S. section, thus it was NOT part of the solution. Furthermore I specifically stated that I made the problem, "I came up with this problem myself because I was trying to explain something that I saw in real life" which would mean that the diagram is correct in explaining the problem.

I posted a question here because I was hoping for some assistance. I gave it my best shot and someone came along and decide to nitpick a little error, rub it in my face, and then say I can't take constructive criticism. Do you have to argue and nitpick everything I say? I'm just a high school physics student, with inane curiosity, who dropped into these forums for help with a complicated and different problem.

By the way, there is another "tiny insignificant" problem...

part b) is impossible. At 35 degrees, the projectile cannot clear the wall. It can only reach a maximum vertical height of approximately 10.14 m up the wall. (Unless it is carried by an updraft!)
Hah-hah. I love the sarcasm there. /sarcasm

Thank you for pointing that out, would you mind telling me if there is a problem with the equations I came up with or with my algebra to find that answer.

Thank you.

Sincerely,
Mator
 
  • #13
754
1
You put the P.S. after the "The attempt at a solution" section, not after the "Homework Statement " section, so it wasn't apparent that the diagram was part of the problem statement.

But okay, you've made your point (finally).

I understand that you are in high school and this is your first attempt at making your own problem; I'm simply trying to point out some things that could help make your problem statement clearer. Had you put the diagram under the problem statement or referred to it in the text of the problem, then there would have been no confusion.

None of my statements were meant to "nitpick" or "rub your face" in anything. I was simply trying to make you see that your problem needed a little polishing.


I'll answer your question concerning part b) in another post...
 
  • #14
11
0
You put the P.S. after the "The attempt at a solution" section, not after the "Homework Statement " section, so it wasn't apparent that the diagram was part of the problem statement.

But okay, you've made your point (finally).

I understand that you are in high school and this is your first attempt at making your own problem; I'm simply trying to point out some things that could help make your problem statement clearer. Had you put the diagram under the problem statement or referred to it in the text of the problem, then there would have been no confusion.

None of my statements were meant to "nitpick" or "rub your face" in anything. I was simply trying to make you see that your problem needed a little polishing.


I'll answer your question concerning part b) in another post...
I understand that you had good intentions, almost all people do, your manner just came across to me in a way that was negative and nit-picking, along with a lot of other pretty words.

I'm not here to argue with you, so I'm not going to attempt to dismantle your position any further.

I understand how you would go about finding that you could not clear the wall by treating the trajectory of the projectile as a straight line at 35 degrees, your statement makes sense. But if what you say is true than I have a few major problems. For one, my problem doesn't accurately represent what I was observing if the projectile cannot clear the wall at the angle I give (the ratio between the distance between the wall and the hill, the height of the hill and the height of the wall and the angle at which the projectile is launched is obviously inaccurate) then I need to change the variables in my problem.

However, since I did indeed get a velocity that means that one the mathematical expressions I generated via the quadratic formula is incorrect. So, what did I do wrong, what step did I mess up at, and how can I fix it?

-Mator
 
  • #15
754
1
I understand that you had good intentions, almost all people do, your manner just came across to me in a way that was negative and nit-picking, along with a lot of other pretty words.
I'm sorry you took my statements that way. You seemed to have gotten irritated with me after my first two posts, but quite honestly, after re-reading them, I don't see how anyone would take my statements as negative and/or nit-picking.

At any rate, I'm happy to help you in any way I can.



Your problem starts when you set [itex]Y_0 + V \sin(q)t + \frac{1}{2}gt^2[/tex] equal to mx

In the equation [itex]Y_0 + V \sin(q)t + \frac{1}{2}gt^2[/tex], Y represents the distance the projectile travels based on it’s starting position [itex](Y_0)[/tex], it’s initial velocity (V), the angle of trajectory (q), gravity (g) and the time it is in flight (t).

In the equation y = mx, y merely represents the height of the wall at a given horizontal offset (x) based on the slope of the wall (m)

Suppose we move the point of launch 2 meters to the left. Now the base of the wall is 10 meters away (horizontally) from the point of launch. The wall still has the equation y = 2.8x and the vertical displacement is still calculated using [itex]Y_0 + V \sin(q)t + \frac{1}{2}gt^2[/tex]
but, obviously the point of impact with the wall (or the height above the wall, if it were possible) would change.


Instead, think of it this way...


Forget about the wall having an equation that describes it; rather consider it’s slope to find the displacement between the point of launch and the top of the wall.

If you do the math, you will see that the top of the wall lies approximately 15.143 meters horizontally from the point of launch ([itex]\frac{106}{7}[/tex] meters, to be exact) and 18 meters away vertically.

So, given that the horizontal velocity is Vcos(q), you can determine how long it will take for the projectile to cover 15.143 meters:

Since x = Vcos(q)t, then t= x/(Vcos(q)), and
t = 15.143/(V cos(q))

Now, use the time t to determine the vertical displacement of the projectile:
[itex]V \sin(q)t + \frac{1}{2}gt^2[/tex]
y = V sin(q) + ½ g t^2
Since we are trying to achieve a (minimum) height of 18 meters in order to clear the wall, we set y = 18

Substitute for t and solve for V.

See what you get then...
 
  • #16
754
1
I forgot to mention...

Using trig, it's easy to see that the top of the wall lies at an angle of

[tex]\tan^{-1}\left( \frac{18}{15.143} \right) = 49.927^\circ[/tex]

above the launch point.
 

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