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Projectile difficult

  1. Jun 1, 2013 #1
    1. The problem statement, all variables and given/known data

    1. Find the maximum angle of projection of a projectile such that its position vector from the origin to the subsequent position of the projectile is always increasing.


    2. Relevant equations



    3. The attempt at a solution
    x(t)=v*cos(A)t, y(t)=v*sin(A)t-(1/2)gt^2. v is the initial velocity, A is the angle from the horizontal.
    then x^2 +y^2 =k^2
    We know that k^2 should always be increasing then we would differentiate it w.r.t to time and put it ≥ 0. But the problem is that i am getting two times for it. now what to do..?
     
  2. jcsd
  3. Jun 1, 2013 #2
    If you get some values of t, that means the magnitude of displacement will be increasing sometimes. You must find a condition when it always increases.
     
  4. Jun 1, 2013 #3
    i have a doubt when i differentiate it wrt to time i assume that theta is constant. can i do that..??
     
  5. Jun 1, 2013 #4
    Does the initial angle of projection change as time goes on?
     
  6. Jun 1, 2013 #5
    but we had to find the maximum angle of projection..
     
  7. Jun 1, 2013 #6
    And?
     
  8. Jun 1, 2013 #7
    only angle of projection for which the position vector always increases..
     
  9. Jun 1, 2013 #8
    You don't have to repeat the problem.
     
  10. Jun 1, 2013 #9
    so is it correct to assume it constant..?
     
  11. Jun 1, 2013 #10
    Answer #4.
     
  12. Jun 1, 2013 #11
    thnx answer is coming sin inverse 1/u
     
  13. Jun 1, 2013 #12
    I do not think this is correct. Show how you got that.
     
  14. Jun 1, 2013 #13
    r^2 = (u^2)t + (g^2)(t^4) - (usinθ)gt^3
    differentiating it and putting it equal to 0 we get
    (gt)^2 -3usinθgt +2u^2 =0
    so t = (3usinθ ± √9u^2sin^2θ-8u^2)/20
    and also this time ≤2usinθ/g (total time of flight)]
    solving this i dont get the answer...i was getting that because of an error.
    now what to do?
     
  15. Jun 1, 2013 #14
    Explain how you got (u^2)t in r^2 = (u^2)t + (g^2)(t^4) - (usinθ)gt^3.
     
  16. Jun 1, 2013 #15
    typing error it is . the actual it is

    (u^2)t^2 + [(g^2)(t^4)/4] - (usinθ)gt^3.
     
  17. Jun 1, 2013 #16
    So how can one ensure that (gt)^2 -3usinθgt +2u^2 is ALWAYS greater than zero?
     
  18. Jun 1, 2013 #17
    if d is less than or equal to zero.
     
  19. Jun 1, 2013 #18
    answer is coming that 8/9≥(sinθ)^2
     
  20. Jun 1, 2013 #19
    is that correct?
     
  21. Jun 1, 2013 #20
    This is not yet the answer. But you are on the right track.
     
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