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Horizontal range of 2 different angles

  1. Jan 27, 2014 #1
    1. The problem statement, all variables and given/known data

    Show that, for a given initial speed, te horizontal range of a projectile is the same for launch angles [itex]45° +\alpha[/itex] and [itex]45° - \alpha[/itex]



    3. The attempt at a solution

    [itex]x1 = vi cos (45° + \alpha)[/itex]
    [itex]x2 = vi cos (45° - \alpha)[/itex]

    tcomplete trajectory = [itex]2 vi sin(45°\pm\alpha)/g[/itex]

    Substituting tcomplete trajectory into x1 and x2:

    [itex]x1[/itex] = [itex]vi^2 sin (2(45°+\alpha))/g[/itex]
    [itex]x2[/itex] = [itex]vi^2 sin (2(45°-\alpha))/g[/itex]
     
  2. jcsd
  3. Jan 27, 2014 #2
    So, are ##x_1## and ##x_2## equal?
     
  4. Jan 27, 2014 #3

    It doesn't appears so.
     
  5. Jan 27, 2014 #4
    Why not? Why can't ## \sin (2(45^\circ + \alpha)) ## be equal to ## \sin (2(45^\circ - \alpha)) ##?
     
  6. Jan 27, 2014 #5

    Let (45 + a) =x and (45 -a) = y
    How does sin(2x) = sin(2y)?
     
  7. Jan 27, 2014 #6
    You are trying to solve a harder problem that way. Stick with the original form.
     
  8. Jan 27, 2014 #7

    The solution in the op is the answer?
     
  9. Jan 27, 2014 #8
    It is very close to the answer. You just need to explain why those two sines are equal. You may want to review the trig identities that you know, or just go straight to the unit circle.
     
  10. Jan 27, 2014 #9

    Ok. I shall further that attempt. It's probably in the form 2cos(x)sin(x)
     
  11. Jan 27, 2014 #10
    One useful identity is $$ \sin \left( \frac \pi 2 + \theta \right) = ? $$
     
  12. Jan 27, 2014 #11
    Solved and Thanks!
     
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