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Projectile motion hitting angle

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data

    You're given a cannon that allows you to fix the initial velocity and the shooting angle of a projectile that should be embedded on the side of a mountain [itex]l[/itex] meters away.

    What is the minimum velocity for you to embed the projectile at an angle of [itex] \frac{\pi}{4}[/itex]?

    3. The attempt at a solution

    I'm not quite sure how to tackle the problem, I do know I have to optimize somehow the velocity but I don't know which expression should I take the derivative of. All help will be greatly appreciated.

    Best,

    M.
     
  2. jcsd
  3. Feb 23, 2014 #2

    haruspex

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    Always start by writing out the equations of motion, creating variables as necessary.
    You don't give a height for the target. If none is given, what height (relative to the cannon) do you think you should assume?
    Of the variables velocity, time, launch angle, x-position, y-position etc., which ones do you think are relevant? Eliminate the others to arrive at one equation with one variable, then optimise it.
     
  4. Feb 23, 2014 #3
    Hey haruspex thanks for replying.
    As you suggested I had indeed written down the equations of motion (I had to because it is a problem with multiple questions with the one I posted as the last one). The thing I'm Confused with is the one you note in your reply, I don't know what height to assume. I guess at the same level as the cannon but don't know how to support that answer. I think the only variables that play a role in the answer are shooting angle and initial velocity, and I also know that for the projectile to hit at [itex]\frac{\pi}{4}[/itex] the components of the velocity i.e. [itex]v_{fy}[/itex] & [itex]v_{fx}[/itex] right before the impact should be the same but still I don't know how to get an answer analytically.
     
  5. Feb 23, 2014 #4

    haruspex

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    No, you need to assume the height doesn't matter. If you were given the range, the height and the arrival angle then you'd have no room for movement. The launch angle and speed would already be determined.
    So, please post some equations. Let the launch be speed v angle theta. What does the range L give you for the flight time? What would the horizontal and vertical velocities be at impact?
     
  6. Feb 23, 2014 #5
    Oh I get it!
    Ok for flight time I got [tex]t_f = \frac{L}{v cos \theta}[/tex] as for velocities the x component is the same as the launch one i.e. [tex]v_{fx} = v cos \theta[/tex] and for the y component [tex]v_{fy} = v sin \theta - \frac{gL}{v cos \theta}[/tex]
     
  7. Feb 23, 2014 #6

    haruspex

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    Right, so what is the impact angle from those?
     
  8. Feb 23, 2014 #7
    The angle is then [tex] \theta_i = tan^{-1}\left( tan \theta - \frac{gL}{(v cos \theta)^2}\right)[/tex]
     
  9. Feb 23, 2014 #8

    haruspex

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    Yes, and that has to be equal to... ?
     
  10. Feb 23, 2014 #9
    Obviously to [itex]\frac{\pi}{4}[/itex]. And then should I take the derivative with respect to v and optimize??
     
  11. Feb 23, 2014 #10

    haruspex

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    Yes, but I suggest going back to your vfx, vfy equations. How can you express a 45 degree downward impact angle in terms of a relationship between those velocities?
     
  12. Feb 23, 2014 #11
    I think something like [tex]\frac{v_{fy}}{v_{fx}} = -1[/tex] should do it, right?
     
  13. Feb 23, 2014 #12

    haruspex

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    Looks good.
     
  14. Feb 23, 2014 #13
    That's some quality help you gave me, I appreciate it!

    Many thanks,

    M.
     
  15. Feb 24, 2014 #14

    haruspex

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    You're welcome. Thank you for an interesting problem.
     
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