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Projectile motion /

  1. Feb 1, 2012 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m and initial velocity v0 is at (x,z)=(0,0) at t=0. The particle should hit the point (x1,z1). What is the angle for which the velocity is minimum? (Constant gravitational field -g in the z direction)

    2. Relevant equations

    v0x = v0*cos(theta)
    v0z = v0*sin(theta)

    3. The attempt at a solution

    I obtain my equations of motion:

    x(t) = t * v0*cos(theta)
    z(t) = t * v0*sin(theta) - (1/2) * g * t^2

    I set x(t=t1) = x1 and solve for the time t1. Then I replace t1 in the equation for z(t=t1)=z1 and solve for v0. Then, I differentiate with respect of theta and set the derivative equal to zero to find a minimum.

    I get:

    1 = 2 * (tan(theta))^2 - 2*z1*tan(theta)/x1 ----------- WRONG

    EDIT :

    2*x1*(sin(θ))^2 - 2*z1*sin(θ)*cos(θ) = x1

    I just don't know how to solve for theta here. It seems to me that I am over complicating things and that there should be an easier way of doing it. I just don't see it ;(.
     
    Last edited: Feb 1, 2012
  2. jcsd
  3. Feb 1, 2012 #2

    Simon Bridge

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    Isn't that a quadratic in [itex]\tan\theta[/itex]?
     
  4. Feb 1, 2012 #3
    Yes. But one gets two solutions. Which one should one choose?
    Both make physical sense to me.
     
  5. Feb 1, 2012 #4
    choose the solution that makes tantheta a real solution.
     
  6. Feb 1, 2012 #5
    Perhaps I am missing something but I get:

    tan(theta) = z1 (+/-) sqrt(z1^2 + 2 * x1^2) / (2 * x1)

    How do I know which one makes it real?

    Thanks.
     
  7. Feb 1, 2012 #6
    depends on the domain of the function. so you need to ask yourself

    does tan(theta)=z1+sqrt(z1^2 + 2x1^2)/2x1 exist

    or does tan(theta)=z1 - sqrt(z1^2 + 2 * x1^2) / (2 * x1) exist
     
  8. Feb 1, 2012 #7
    I think I get it. The solution with a minus leads to a negative for tan(theta). But that would lead to negative thetas right?
     
  9. Feb 1, 2012 #8
    So, the solution should be real for both
     
  10. Feb 1, 2012 #9
    but the negative solution doesn't make sense. do you agree?
     
  11. Feb 1, 2012 #10
    Yes. But I made a mistake ;(. I forgot about a sec(theta)^2 factor when I differentiated v0.
     
  12. Feb 1, 2012 #11
    Now I get:

    2*x1*(sin(θ))^2 - 2*z1*sin(θ)*cos(θ) = x1

    after differentiating and setting equal to zero.
     
  13. Feb 1, 2012 #12

    Simon Bridge

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    You need trig identities there. (Tidier if you put k=x1/z1 too.)
    I think you are doing it the hard way - did you try doing the geometry of the v-t graphs?
    Plotting different-speed trajectories to find the trends?

    [edit: I take that bac - trig IDs are easier]
     
    Last edited: Feb 1, 2012
  14. Feb 1, 2012 #13
    Thank you for replying. But how would looking at the v-t graph help me find the angle for minimum initial velocity? That would just give me information about the magnitude of the velocity at any time. I'm only concerned about the initial velocity and which angle makes it a minimum so that it hits a point (x1,z1).
     
  15. Feb 1, 2012 #14

    Simon Bridge

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    I've done a similar problem - the v-t geometry showed a shortcut.
    However - from where you are, the trig IDs will do what you want.
     
  16. Feb 1, 2012 #15
    "A particle of mass m and initial velocity v0 is at (x,z)=(0,0) at t=0. The particle should hit the point (x1,z1). What is the angle for which the velocity is minimum? (Constant gravitational field -g in the z direction"

    In my opinion, the best way to approach a physics problem is understanding what we are asked. So, What exactly do they mean by velocity? Initial velocity or what type of velocity?
     
  17. Feb 2, 2012 #16

    Simon Bridge

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    We have been provided with the context for the question: projectile motion.
    I'll agree that it is important to understand the problem.

    basically the problem is to find a parabolic trajectory through two points which also satisfies the constraint that the instantaneous speed at the origin is a minimum.

    IT is possible to get an idea where the solution lies by examining this picture. For instance, the angle wanted must be bigger than A=arctan(z1/x1) ... we can also consider the difference if this angle is greater or less than 45deg, or what happens if z1 < 0. All builds a picture. If we point the gun just above A, we need a fast speed to hit the target. Similarly, a very indirect path will also need a fast speed. So the answer is in between.

    Thing is - OP has already gone a long way by brute force, and only need the trig IDs for his final relation to yield fruit.
     
  18. Feb 2, 2012 #17
    I'm sorry. It should be angle for minimum velocity.
     
  19. Feb 2, 2012 #18
    I can understand a question that may ask about optimizing distance, but optimizing velocity is funny, because Then you would need to refer to a direction and how the velocity relates to optimization. For example, If they said find the angle of minimum velocity in terms of the x direction, then that's easy. The angle is 90 degrees since the projectile would be thrown completely upward and velocity in the x direction would be zero. See what I'm thinking here?
     
  20. Feb 2, 2012 #19
    But the initial velocity doesn't depend on the trajectory, unless I'm wrong. The initial velocity is fixed and everything else is calculated thereafter.
     
  21. Feb 2, 2012 #20

    Simon Bridge

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    @Rayquesto - nope: that is not correct.
    The target is fixed. Speed and angle are up for grabs.
    Which is actually how projectile problems work IRL. The problem, in essence, is to find the cheapest firing solution.
    That's right - everything is all mixed up.
     
    Last edited: Feb 2, 2012
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