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Find unknown angle of Projectile

  1. Mar 2, 2015 #1

    h.s

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    I don't need the solution solved for me, but I need help with the method to find the solution.
    1. The problem statement, all variables and given/known data

    A projectile is launched from an unknown angle and expected to hit a target at a lower altitude.
    The values given are the initial velocity, and the distance of the target from the launcher.

    2. Relevant equations


    3. The attempt at a solution
    I've tried solving for time, angle, or final velocity but they each need another unknown variable. Any insight would be helpful.
     
  2. jcsd
  3. Mar 2, 2015 #2

    PeroK

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    A trajectory can be described as a parabola of the form ##y = ax^2 + bx + c##, where usually ##c = 0## and ##a, b## are functions of ##v, \theta##.

    If you derive this equation that would be a good start.
     
  4. Mar 2, 2015 #3

    h.s

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    i don't have θ. would a be the initial velocity or average velocity? because I can't solve for average velocity
     
  5. Mar 2, 2015 #4

    PeroK

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    The parabolic equation is a general equation that doesn't require you to know the numerical value of any parameter. Do you understand the concept of working with unknowns as parameters, such as ##v, \theta, x \ \ and \ \ y##?
     
  6. Mar 2, 2015 #5

    h.s

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    I'm still at a loss. Suppose I apply this equation. The line would pass through two coordinates (0, 1.13) at launch, then at (4.382, 0) when it hits the ground. How would I solve for θ?
     
  7. Mar 2, 2015 #6

    PeroK

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    You'll have one equation with one unknown. Solving for ##\theta## should be relatively simple if you can do the maths to get there.
     
  8. Mar 2, 2015 #7

    h.s

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    ... I'll have one equation with two unknowns; I still don't have a and b..
     
  9. Mar 2, 2015 #8

    PeroK

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    a and b will be functions of v (which you know) and ##\theta##

    I'll give you a start:

    ##x = vtcos \theta##

    ##y = ?##

    Then, try to eliminate t. It's over to you to do some maths, I'm afraid!
     
  10. Mar 2, 2015 #9

    h.s

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    What does that mean?
     
  11. Mar 2, 2015 #10

    h.s

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    i know x=vtcosθ and y=vtsinθ-1/2gt^2, but my problem is that I don't know how to go further without knowing t, nor θ
     
  12. Mar 2, 2015 #11

    PeroK

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    ##x=vtcos\theta \ \Rightarrow \ t = \frac{x}{vcos\theta}##
     
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