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Projectile Motion with initial velocity

  1. Sep 22, 2014 #1
    1. The problem statement, all variables and given/known data
    A projectile with initial velocity u making an angle A with the horizontal. Find the time t when the velocity of the projectile becomes perpendicular to its original velocity.


    2. Relevant equations
    v = u + at
    s = ut + 1/2at^2
    v^2 - u^2 = 2as




    3. The attempt at a solution
    I have an intuitive idea of how to solve it. When the velocity becomes perpendicular to the original velocity, their dot product will be zero. I don't know if this is the right way. If this the right way, I don't know how to solve for it.
     
  2. jcsd
  3. Sep 22, 2014 #2

    Orodruin

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    That indeed sounds like a good approach. So how can you realize your idea? What is the velocity vector at a given time?
     
  4. Sep 22, 2014 #3
    It's the vector sum of v(y) = u sinA - gt and v(x) = u cosA
     
  5. Sep 22, 2014 #4

    Orodruin

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    Indeed. So what is the inner product of that vector at ##t=0## with that vector at ##t=T##?
     
  6. Sep 22, 2014 #5
    That's were the problem is. I got it as the square root of v^2 - 2 u g t sinA + g^2 t^2. I don't know what to do further.
     
  7. Sep 22, 2014 #6

    Orodruin

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    That looks as the square root of the actual inner product (assuming the v^2 should be u^2), but it should be zero anyway so this does not matter much.

    So, do you know how to solve for the roots of a second order polynomial?
     
  8. Sep 22, 2014 #7
    Through the Discriminant?
     
  9. Sep 22, 2014 #8

    Orodruin

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    The discriminant gives you some information on the nature of the roots. Let us take something more hands on: If
    $$
    x^2 + px + q = 0,
    $$
    what is ##x##?
     
  10. Sep 22, 2014 #9

    Orodruin

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    Sorry, I am not thinking straight. The expression you have is not the inner product but the speed at time t. How is the inner product of two vectors defined?
     
  11. Sep 22, 2014 #10
    Mod note: I have deleted the content of this post, as it was a complete solution.
     
    Last edited by a moderator: Sep 22, 2014
  12. Sep 22, 2014 #11
    just realized i have made the mistake of not making B negative (since the velocity should be downwards)

    so the expression should be t=(u/gsinA)
     
  13. Sep 22, 2014 #12

    Orodruin

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    This expression is not correct as it gives A+B = pi/2 which is not equivalent to orthogonality. Also, note that forum rules forbids posting full solutions until the OP has solved the problem.
     
  14. Sep 22, 2014 #13
    doesn't that make the angle between the velocities orthogonal?
     
  15. Sep 22, 2014 #14
    vector approach:
    u.v=0

    (ucosA i + usinA j).(vcos(-B) i + vsin(-B) j )=0

    (ucosA i + usinA j).(ucos(A) i + (usinA-gt) j )=0
     
  16. Sep 22, 2014 #15

    Orodruin

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    No, in order to be orthogonal you would need A-B = pi/2. That B is negative is being taken care of automatically by the vector component.
     
  17. Sep 22, 2014 #16
    yup I realized that mistake
     
  18. Sep 23, 2014 #17
    I used this but got an expression I couldn't solve.
     
  19. Sep 23, 2014 #18
    I used this, but could not solve the expression I got.
     
  20. Sep 23, 2014 #19

    Orodruin

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    If you do the inner product equal to zero approach you should get a linear expression in t. It should be a simple matter of solving this.

    Note that what you had earlier was not the inner product of the velocity at t and that at time 0, but the speed at time t. That will never be zero unless the horizontal component is.
     
  21. Sep 23, 2014 #20
    (ucosA i + usinA j).(ucos(A) i + (usinA-gt) j )=0

    I have to this, right?
     
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