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Problem on projectile

  1. Jul 10, 2013 #1
    See the attachment.

    The right answer is θ=60 degrees

    but i am getting sinθ = ±√3.
     

    Attached Files:

  2. jcsd
  3. Jul 10, 2013 #2
    Your expression for ##v_y## is wrong.

    Also, check your last step when you plug 3 in 4.
     
    Last edited: Jul 10, 2013
  4. Jul 10, 2013 #3
    I don't agree with your very first equation. It is not true that
    [tex]H= \frac{u^{2} \sin^{2}( \theta)}{2g}.[/tex]
    Conservation of Energy requires, instead, that
    [tex]mgH+ \frac{m u^{2} \cos^{2}( \theta)}{2}= \frac{mu^{2}}{2},[/tex]
    comparing the peak to the starting-point, or
    [tex]gH+ \frac{u^{2} \cos^{2}( \theta)}{2}= \frac{u^{2}}{2}.[/tex]
    Then you can also write
    [tex] \frac{gH}{2}+ \frac{5 u^{2} \cos^{2}( \theta)}{4}= \frac{u^{2}}{2},[/tex]
    comparing the energies at the half-way point to the starting-point.
     
  5. Jul 10, 2013 #4
    First things first.

    How is my expression for vy wrong?

    I used v2-u2 = 2as
     
  6. Jul 10, 2013 #5
    What are the directions for velocity in vertical direction and acceleration?
     
  7. Jul 10, 2013 #6

    haruspex

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    Doesn't that reduce to the same equation?
     
  8. Jul 10, 2013 #7
    You're quite right. Chalk that one up to needing to do one step at a time. Also, for the half-way point, it doesn't reduce the same way.
     
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