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Projectile Motion, dying here

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A projectile is fired into the air from the top of a 200 m cliff above a valley as shown in the figure. Its initial velocity is 60 m/s above the horizontal. Where does the projectile land?




    2. Relevant equations

    x=60cos60 t

    set the next equation equal to zero and solve for t:

    -200=60sin60t-1/2(-9.81)(t)^2

    3. The attempt at a solution

    I solved for t so I could find horizontal displacement, ended up with the following quadratic but end up with a negative number inside square root.
    4.9t^2+52t+200=0 I know my mistake is probably obvious but I cant see it, help appreciated.
     
  2. jcsd
  3. Sep 14, 2009 #2

    rl.bhat

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    -200=60sin60t-1/2(-9.81)(t)^2
    In the equation you have to use only one negative.
     
  4. Sep 14, 2009 #3

    ideasrule

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    The equation is displacement=v0t+(1/2)at^2. You have to choose one direction as positive and write all quantities according to that convention. If you do that, you'll end up with only one negative, as rl said.
     
  5. Sep 14, 2009 #4
    But isn't gravity negative? So don't I plug a negative value here???
     
  6. Sep 14, 2009 #5
    Appreciate it
     
  7. Sep 14, 2009 #6

    ideasrule

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    No, you always use the equation:

    displacement=v0t+(1/2)at^2

    If you choose "up" as positive, then displacement would be -200, which you have. v0 would be vsin(60), which you also have. a would be -9.8 m/s^2, since gravity acts downwards. Leave the positive sign in front of 1/2 alone; that can never change.
     
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