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Trajectory Problem

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data
    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).


    2. Relevant equations
    (vsin(a)+sqrt(vsin(a)^2+2(9.8)(41))/9.8


    3. The attempt at a solution
    STill attempting.
     
  2. jcsd
  3. Jul 19, 2010 #2

    Redbelly98

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    Welcome to Physics Forums.
    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.
     
  4. Jul 19, 2010 #3

    kuruman

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    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?
     
  5. Jul 19, 2010 #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
     
  6. Jul 19, 2010 #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).
     
    Last edited: Jul 19, 2010
  7. Jul 19, 2010 #6

    kuruman

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    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.
    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?
     
  8. Jul 19, 2010 #7

    kuruman

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    What possessed you to do this? The expression sqrt(2x/g) has dimensions of time, not velocity. You just can't do this.
     
  9. Jul 19, 2010 #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. :|
     
  10. Jul 19, 2010 #9

    kuruman

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    Which "distance" would that be?
     
  11. Jul 19, 2010 #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. :|
     
  12. Jul 19, 2010 #11

    kuruman

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    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?
     
  13. Jul 19, 2010 #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.
     
  14. Jul 19, 2010 #13

    kuruman

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    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.
     
  15. Jul 19, 2010 #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.
     
  16. Jul 19, 2010 #15

    kuruman

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    You are not correct saying that. v0 is indeed the muzzle velocity. It is zero only before the projectile is fired.
     
  17. Jul 19, 2010 #16
    Okay, so then V would in fact be f(v) and that's the number I will be plugging.
     
  18. Jul 19, 2010 #17

    kuruman

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    I don't know what you mean. Here v is the muzzle velocity, the independent variable. What does f(v) represent do you think?
     
  19. Jul 19, 2010 #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? :|
     
  20. Jul 19, 2010 #19

    kuruman

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    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.
     
  21. Jul 19, 2010 #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...
     
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