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Kinematics, with two unknowns

  1. Aug 8, 2011 #1
    1. The problem statement, all variables and given/known data

    A projectile is fired at an angle 60 degrees above the horizontal with a speed of 60m/s. A 0.5 second later, another projectile fired at the same speed but at a different angle. The two projectiles collide at some point (x,y). Find the angle the second projectile was fired and the coordinate (x,y) they collide.

    3. The attempt at a solution

    All subscripts "1" refer to the first projectile and the "2" refer to the second projectile.

    [tex]y_1 (t) = \frac{-gt_1^2}{2} + 30\sqrt{3}t_1[/tex]

    [tex]y_2 (t) = \frac{-g(t_1 + \frac{1}{2})^2}{2} + 60(t_1 + \frac{1}{2})\sin\alpha[/tex]

    [tex]x_1 (t) = 30t_1[/tex]
    [tex]x_2 (t) = 60(t_1 + \frac{1}{2})\cos\alpha[/tex]

    Naturally I set them equal to each other to find that (x,y) coordinate

    [tex]y_1 (t) = y_2 (t)[/tex]

    [tex]30\sqrt{3} t_1 = \frac{-g}{2}(t_1 + \frac{1}{4}) + 60(t_1 + \frac{1}{2})\sin\alpha[/tex]

    (2) [tex](30\sqrt{3} + \frac{g}{2})t_1 + \frac{g}{8} = 60(t_1 + \frac{1}{2})\sin\alpha[/tex]

    [tex]x_1 (t) = x_2 (t)[/tex]

    (1) [tex]30t_1 = 60(t_1 + \frac{1}{2})\cos\alpha[/tex]

    So from Euclid's elements, I can now divide (2) by (1)

    [tex]\frac{(30\sqrt{3} + \frac{g}{2})t_1 + \frac{g}{8}}{30t_1} = \tan\alpha[/tex]

    [tex]\sqrt{3} + \frac{g}{60} + \frac{g}{240t_1} = \tan\alpha[/tex]

    Now here is the problem, what is t1? Will all values (other than 0) of t1 work?

    I tried to go to back to my other basic kinematics equation, but I don't want a particular t1, I need a general one?
    Last edited: Aug 8, 2011
  2. jcsd
  3. Aug 8, 2011 #2
    You have two equations and two unknown. So it can be solved. Just solve [itex]t[/itex] in term of [itex]\alpha[/itex] from one equation and substitute it into the second equations.

    BTW, you original equations are wrong. the trajectory for [itex]x_2[/itex] and [itex]y_2[/itex] should have [itex]t_1 - 0.5[/itex] instead of [itex]t_1+0.5[/itex]. The travel time for the second projectile is less than the first one.
  4. Aug 8, 2011 #3
    first, i think u have put it wrong, the time of 2nd projectile, it has been fired 1/2 sec later, so it will take less time to collide or travel the same distance with the first. so the time will be t1-1/2. 2nd-ly u can always find t1 in terms of cos alfa from the eqn. x1(t1) = x2(t2). replace t1 in terms of cos alfa in the eqn. u got by deviding eqn. 1 with 2. there will be only one unknown, alfa. it can be solved with a bit of mathematics. i have not worked it out myself, so i don't know how cumbersome it may be.
  5. Aug 8, 2011 #4
    So I got

    [tex]\sqrt{3} + \frac{g}{240t_1} - \frac{g}{60} = \tan\alpha[/tex]

    From x_1 = x_2

    [tex]t_1 = \cos\alpha[/tex]

    [tex]\sqrt{3} + \frac{g}{240\cos\alpha} - \frac{g}{60} = \tan\alpha[/tex]

    Any know any trig identities to simplify this?
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