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Projectile Motion

  1. Dec 5, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball was projected at an angle A to the horizontal. One second later another ball was projected from the same point at an angle B to the horizontal. One second after the second ball was released, the two balls collided. Find the speed of projection for the two balls.


    2. Relevant equations
    s = ut + 1/2 at2


    3. The attempt at a solution
    In x-direction:
    sA = ua cosA t
    sB = ub cosB (t-1)

    In y-direction:
    sA = ua sinA t -1/2 gt2
    sB = ub sinB (t-1) -1/2 g(t-1)2

    When they collide the distances they travelled in the x and y directions are equal to each other at t=2. I tried to solve for ua and ub but got stuck.
     
  2. jcsd
  3. Dec 5, 2009 #2

    rl.bhat

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    Homework Helper

    Hi whiteman, welcome to PF.
    When the balls collide, ball A is in the air for 2 s and ball B is in the air for 1 s.
    For them x distance is the same. So
    2*Ua*cosA = Ub*cosB -----(1)
    For y
    2*Ua*sinA - 1/2*g*(2)^2 = Ub*sinB - 1/2*g -------(2)
    From eq.(1), find the expression for Ub and substitute it in equation (2) and solve for Ua.
     
  4. Dec 5, 2009 #3
    Hi, thanks for the welcome.
    Is Ua and Ub meant to expressed in terms of cos/sin of A/B or are they just numbers?
    I got Ua = 3gcosB/4sin(A+B) and Ub = 3gcosA/2sin(A+B).
    Is this right or am I going wrong?
     
  5. Dec 5, 2009 #4

    rl.bhat

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    Homework Helper

    In the expression of Ua and Ub sin(A-B) should be there instead of sin(A+B)
     
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