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Help with projectile motion, finding x

  1. Oct 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A balloon is attached to a 3m high pole. On the ground is a mortar that fires at a velocity of 7m/s at an angle of 55degrees. How far away does the mortar have to be to hit the balloon?

    2. Relevant equations
    y=tan(theta)x-g/(2v0cos2(theta)x2



    3. The attempt at a solution
    I rearranged to make a quadratic equation from the equation above and made

    g/(2v0cos2(theta)x2 - tan(theta)x+y=0

    -9.81m/s2/(2*(7m/s)2(cos2(55deg)x2 - tan(55)x+3m

    and ended up with 1.58m and -6.34m

    I understand that there will be two solutions because the projectile can hit the balloon on the way up and the way down but i dont think these answers are correct.
     
  2. jcsd
  3. Oct 2, 2014 #2

    vela

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    You seem to have a sign error. However, with the given numbers, I find there's no solution.
     
  4. Oct 2, 2014 #3
    in my lecture the correct answer was something like 1.9 or something and i forget the other one but when i plugged my answer 1.58 back into y=tan(theta)x-g/(2vcos^2(theta)x^2 it comes out as the right y value - 3m.

    is there a chance im using the completely wrong equation for this?
     
  5. Oct 2, 2014 #4

    vela

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    You're using the right equation. Here's a plot of the trajectory with the given numbers. The projectile never reaches a height of 3 m.
     

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  6. Oct 2, 2014 #5

    vela

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    When I plug in x=1.58 m, I get y=1.50 m.
     
  7. Oct 2, 2014 #6
    I feel a little lost then. so for a projectile firing at 7m/s at 55 degrees trying to hit a balloon 3m in the air, it will never actually hit it?
     
  8. Oct 2, 2014 #7

    vela

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    Right. It's not going fast enough to reach a height of 3 m when it's aimed at 55 degrees off the horizontal. In fact, even if you shot it straight up, it wouldn't reach that height.
     
  9. Oct 2, 2014 #8
    im following now. max height is 1.68m if i worked it out correctly so perhaps my notes are wrong. I'll compare notes with a friend and post again. thanks for your help so far
     
  10. Oct 2, 2014 #9

    vela

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    I'm not sure how you got that number. There's something weird going on with your calculations.
     
  11. Oct 2, 2014 #10
    for max height i used

    v0^2sin^2(theta)/2g

    72*sin2(55) / 2 * 9.81 = 1.676
     
  12. Oct 2, 2014 #11

    vela

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    Ah, okay, I thought you were referring to shooting the projectile straight up.
     
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