Solving Trajectory Problem with 2 Variables (V,A)

  • Thread starter Sciencerob
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In summary: After it is projected, it is no longer zero. And the forumula you have is a general one, true for any projectile launched at an angle of theta above the horizontal, with muzzle velocity v0. If you have a particular projectile in mind, with a particular angle of projection, and a particular initial height, then you can use the formula to calculate the flight time, and then the range (as I showed you before), and then you can plot that range vs the muzzle velocity, to get a straight line. Is that what you are after?In summary, the conversation discusses a group project involving projectile motion, specifically finding the relationship between muzzle velocity and range. The initial equation given by the group
  • #1
Sciencerob
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Homework Statement


I was attempting to finish a group project when I noticed certain elements lacking in it. I am attempting to graph the below equation but I can't get it to work. It has two variables, V(Velocity) and A(time in this case). So I am trying to figure out how I can graph this when variable will remain when I have put in a time for F(a).


Homework Equations


(vsin(a)+sqrt(vsin(a)^2+2(9.8)(41))/9.8


The Attempt at a Solution


STill attempting.
 
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  • #2
Welcome to Physics Forums.
Sciencerob said:

Homework Equations


(vsin(a)+sqrt(vsin(a)^2+2(9.8)(41))/9.8
This is not an equation, it is an expression. An equation must have an "=" sign in it. There is no way to graph what you have written here.

Also, you seem to missing a ")" somewhere.
 
  • #3
This looks like the solution of the quadratic equation that gives you the specific time interval at which a projectile, launched at angle a above the horizontal, reaches a height of 41 m below the launching point. Here v is the independent variable and t is the dependent variable. You can certainly plot t vs v, can you not? Just pick different vees and calculate the corresponding tees. Is that not what you want?
 
  • #4
Sorry let me clarify:

This is a standard trajectory problem. A cannon on a fortress wall is at a 30 degree tilt. The equation I have developed from what my group gave me is this:

F(x)= 44.1+Vsin(theta)(x/(Vcos(theta)))-1/2g(x/Vcos(theta))^2 where V is muzzle velocity X is distance and G is -9.8.

Unfortunately they left me(my group who told me the equation was good to go and i just needed to make a graph and turn it in) with MV and X. Obviously the standard formula is V[sub zero] is D/T. As Distance is the variable in question and I can't use time. So I am racking my brain to put Velocity in terms besides those two but I don't think I can do it. I might have to rework the entire formula if I am correct.

So if anyone has any ideas let me know. I'm going to rework the equation, probably from scratch. X_X
 
  • #5
All I need is the equation to relate muzzle velocity to distance so I can graph it. I am thinking maybe instead of calculating Muzzle velocity from distance maybe I should get distance from muzzle velocity?Update:

44.1+(x/((sqrt(2X))/-9.8))sin(30)(x/(x/((sqrt(2X))/-9.8))cos(30))-1/2*-9.8(x/(x/((sqrt(2X))/-9.8))cos(30))^2

I replaced V with the SQRT of 2x/G(acceleration).
 
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  • #6
Sciencerob said:
Obviously the standard formula is V[sub zero] is D/T.
That is not obvious at all. It is obvious only when you have motion in a straight line under zero acceleration. This is not the case here.
As Distance is the variable in question and I can't use time. So I am racking my brain to put Velocity in terms besides those two but I don't think I can do it. I might have to rework the entire formula if I am correct.

So if anyone has any ideas let me know. I'm going to rework the equation, probably from scratch. X_X
The equation is "good to go" if you know the muzzle velocity and the angle of projection. Do you know them? If not, do you know the range and maximum height? More generally, what variables do you know?
 
  • #7
Sciencerob said:
I replaced V with the SQRT of 2x/G(acceleration).
What possessed you to do this? The expression sqrt(2x/g) has dimensions of time, not velocity. You just can't do this.
 
  • #8
I am trying desperatly to put everything in terms of one variable. :(

I do know the height of the cannon on the fortress is 44.1m. Angle theta is 30 degrees. That is all the information I have. I am trying to make a function that can give me muzzle velocity given a distance.

Edit: THinking about it you are right. Time is not even close to velocity. :|
 
  • #9
Sciencerob said:
I am trying to make a function that can give me muzzle velocity given a distance.
Which "distance" would that be?
 
  • #10
There is a theoretical ship out there. I need to determine the velocity needed to hit the ship. The distance is from the fortress(and thus cannon) to the ship. Also it would be the range as I understand it.

And I relooked at the formulas. Given this type of problem, Wouldn't (x/((sqrt(2X))/9.8)) still give you the velocity? That is distance over time after all. Maybe the whole this is butchered... I just don't get why my graph doesn't take a parabola shape, it has the necessary exponents. :|
 
  • #11
OK, so you are looking for the muzzle velocity. So you want to find muzzle velocity as a function of range. Is that it? Or do you have a number for the range and you are looking for a muzzle velocity to match it?
 
  • #12
Yup, that's it Kuru. I need MV as a function of range. The range is variable so I'm not looking for any exact number really. I just need the function so I can show the relationship on a graph. I just don't get where I went wrong. I'm not sure if the equation my group gave me is inadequate or if I butchered it all on my own.
 
  • #13
OK then. The very initial expression that you had, namely

[tex]t_f=\frac{v_0Sin \theta +\sqrt{v^{2}_{0}Sin^2 \theta+2gh}}{g}[/tex]

gives you the time of flight. Multiply this time by the horizontal component of the velocity and you will have the range. So pick yourself a bunch of muzzle velocities and calculate a bunch of ranges. Then plot the results.
 
  • #14
So am I correct is saying V sub-zero is in fact zero? I was initially thinking that V sub zero would be the muzzle velocity.
 
  • #15
Sciencerob said:
So am I correct is saying V sub-zero is in fact zero? I was initially thinking that V sub zero would be the muzzle velocity.
You are not correct saying that. v0 is indeed the muzzle velocity. It is zero only before the projectile is fired.
 
  • #16
Okay, so then V would in fact be f(v) and that's the number I will be plugging.
 
  • #17
Sciencerob said:
Okay, so then V would in fact be f(v) and that's the number I will be plugging.
I don't know what you mean. Here v is the muzzle velocity, the independent variable. What does f(v) represent do you think?
 
  • #18
I mean that if I were to set it up to graph on a calculator or excel that V is the variable I would be inserting the value for. For the graphing program I have on my computer I have done this so far but I think I have made an error, still in the process of trying to fix it.

F(x)=(x*Sin(30)+Sqrt(x^2Sin^2(30)+2(44.1)(9.8))/9.8)*x

I am mistaken in this? :|
 
  • #19
Sciencerob said:
F(x)=(x*Sin(30)+Sqrt(x^2Sin^2(30)+2(44.1)(9.8))/9.8)*x

In the above expression, x represents the muzzle velocity and F(x) is supposed to represent the range. I said "multiply the time of flight by the horizontal component of the initial velocity" to get the range. You didn't do that.
 
  • #20
Well, I thought the first part I had gotten all the variables correct. (x*Sin(30)+Sqrt(x^2Sin^2(30)+2(44.1)(9.8))/9.8) represents the equation you have me. Then multiplying by the X should give the range... unless I have to separate the horizontal component of initial velocity prior to multiplying it...
 
  • #21
Sciencerob said:
Well, I thought the first part I had gotten all the variables correct. (x*Sin(30)+Sqrt(x^2Sin^2(30)+2(44.1)(9.8))/9.8) represents the equation you have me.
As Redbelly98 said this is not even an equation.

Then multiplying by the X should give the range... unless I have to separate the horizontal component of initial velocity prior to multiplying it...
Is "the X" the horizontal component of the velocity? What does "the X" represent?
 
  • #22
The X represents velocity which could is probably moving both in the Y direction and the X direction. I suppose what I should be doing for the last part is V(subx) equals Vsubzero Cos(Theta) to give me X velocity. (I know its not an equation I meant function).

So is F(x)= (x*Sin(30)+Sqrt(x^2Sin^2(30)+2(44.1)(9.8))/9.8)*xcos30 correct?
 
  • #23
X represents the muzzle speed. Is the magnitude of the initial velocity vector and has no direction. It is usually represented by v0 not X. My friendly suggestion to you is to stop thinking in terms of variables that your plotting calculator requires and start thinking in terms of symbols that represent physical variables.

Anyway, I think you are ready to start graphing.
 
  • #24
Alright. Thank you. I do seem to be having some trouble using my Ti-89 though. For some reason when I input it I can't get the sin Squared to format. It tells me that I'm missing a parentheses when I do that. Any possible help with this? :|

Nevermind Found out what to do.

Just to make sure, I put in 500 as the velocity and got out 6.891 as the range. Does that seem plausible? It seems a bit small to me.
 
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  • #25
I have zero experience with the TI-89, however if you use sine*sine instead of trying to square it, it should work.
 
  • #26
I just put in 500 and then 1000 as different variables and they yielded the same answer...

Checking the format the only thing that looks like it might be out of place is that I have it Sin(30)^2 rather than Sin^2(30). It wouldn't take it any other way. I did put gravity as a negative force though. Should it be positive in this case? I think it should be negative considering the trajectory is going upward.
 
  • #27
Sciencerob said:
I just put in 500 and then 1000 as different variables and they yielded the same answer...

Checking the format the only thing that looks like it might be out of place is that I have it Sin(30)^2 rather than Sin^2(30). It wouldn't take it any other way. I did put gravity as a negative force though. Should it be positive in this case? I think it should be negative considering the trajectory is going upward.
The symbol "g" stands for "the magnitude of the acceleration of gravity" and is a positive number equal to 9.8 m/s2. As for your sine problems, I offered you a suggestion, use it or not as you see fit.
 
  • #28
I did use your sin*sin format and I removed the negatives from the gravity. Still getting the same results. I've doubled and triple checked, I did put in the right variables all the way through. I keep getting 6.88337
 
  • #29
Sciencerob said:
I did use your sin*sin format and I removed the negatives from the gravity. Still getting the same results. I've doubled and triple checked, I did put in the right variables all the way through. I keep getting 6.88337
What are the "same results" that you are getting? Say you pick a muzzle speed of 300 m/s. What range do you get?
 
  • #30
When input 300 I get 6.86804. Anything above 70 starts to level off. I've done the following:

10 -> 3.332
20 -> 4.9
30 -> 5.71
40 ->6.12
70 ->6.599
 
  • #31
It looks like you are making a big mistake in your inputs, either the variables or the formula itself, probably the formula. I used Excel and here are a couple of benchmarks

v0 = 150 m/s, R = 2057 m
v0 = 300 m/s, R = 8023 m

I cannot help you any more. It is between you and your TI-89.
 
  • #32
Alright, I've moved to a computer with excel. If anyone would mind helping me out here (sorry for the trouble).

=(500*Sin(30)+Sqrt(500^2*Sin(30)*Sin(30)+2(44.1)(9.8)/9.8))*500cos(30)

is what I put in and it spits back out at me that I am doing something wrong..
 
  • #33
First off you need to put asterisks to denote multiplication. This 2(44.1)(9 .8) won't work.
Secondly, you need to put the angle in radians, not degrees.
 
  • #34
Almost there... I got it to work but when I put in 300 for Velocity I got a number around 40,000.

=(300*SIN(0.5236)+SQRT(300^2*SIN( 0.5236)*SIN(0.5236)+2*(44.1)*(9.8))/9.8)*300*COS( 0.5236)

That is my formula.
 
  • #35
Well, I do hope I can get a response quick before my school's library closes. If not Thanks for your help.
 
<h2>1. How do you calculate the trajectory of an object with 2 variables, velocity and acceleration?</h2><p>To calculate the trajectory of an object with 2 variables, velocity (V) and acceleration (A), you can use the following formula: <strong>Trajectory = V * t + 1/2 * A * t^2</strong>, where t represents time. This formula takes into account the initial velocity and the constant acceleration of the object.</p><h2>2. Can the trajectory of an object with 2 variables be affected by external factors?</h2><p>Yes, the trajectory of an object with 2 variables can be affected by external factors such as air resistance, wind, and friction. These factors can alter the velocity and acceleration of the object, thus changing its trajectory.</p><h2>3. What is the difference between displacement and trajectory?</h2><p>Displacement refers to the overall change in position of an object, while trajectory refers to the path that the object takes in space. In other words, displacement is the distance between the initial and final position, while trajectory is the actual route the object takes to get from one point to another.</p><h2>4. How can you use the equations of motion to solve trajectory problems with 2 variables?</h2><p>The equations of motion, such as the one mentioned in the first question, can be used to solve trajectory problems with 2 variables by plugging in the known values for velocity, acceleration, and time. By rearranging the equation, you can solve for any missing variable.</p><h2>5. Is it possible to have a negative trajectory?</h2><p>Yes, it is possible to have a negative trajectory. This would occur if the object is moving in the opposite direction of the positive direction, or if it experiences a negative acceleration. In this case, the trajectory would have a downward slope on a graph. </p>

1. How do you calculate the trajectory of an object with 2 variables, velocity and acceleration?

To calculate the trajectory of an object with 2 variables, velocity (V) and acceleration (A), you can use the following formula: Trajectory = V * t + 1/2 * A * t^2, where t represents time. This formula takes into account the initial velocity and the constant acceleration of the object.

2. Can the trajectory of an object with 2 variables be affected by external factors?

Yes, the trajectory of an object with 2 variables can be affected by external factors such as air resistance, wind, and friction. These factors can alter the velocity and acceleration of the object, thus changing its trajectory.

3. What is the difference between displacement and trajectory?

Displacement refers to the overall change in position of an object, while trajectory refers to the path that the object takes in space. In other words, displacement is the distance between the initial and final position, while trajectory is the actual route the object takes to get from one point to another.

4. How can you use the equations of motion to solve trajectory problems with 2 variables?

The equations of motion, such as the one mentioned in the first question, can be used to solve trajectory problems with 2 variables by plugging in the known values for velocity, acceleration, and time. By rearranging the equation, you can solve for any missing variable.

5. Is it possible to have a negative trajectory?

Yes, it is possible to have a negative trajectory. This would occur if the object is moving in the opposite direction of the positive direction, or if it experiences a negative acceleration. In this case, the trajectory would have a downward slope on a graph.

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