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Projectile doubt.

  1. May 16, 2007 #1
    Projectile doubt.......

    CHOOSE THE CORRECT ANSWER AND EXPLAIN.

    What is the nature of proportionality between the muzzle angle and the horizontal distance traveled by a projectile?

    a) Directly proportional.
    b) Inversely proportional.
    c) None of the above.


    (:biggrin: who says learning physics is boring?:tongue2: )
     
    Last edited: May 16, 2007
  2. jcsd
  3. May 16, 2007 #2

    Doc Al

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    Staff: Mentor

    Well, what do you think?
     
  4. May 17, 2007 #3
    Well think about this. The bigger the angle, the higher the projectile goes. The higher it goes, the longer time it takes to reach the ground. The longer it takes to reach the ground,...
    You don't even need formulas for this.
     
  5. May 17, 2007 #4

    Doc Al

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    Staff: Mentor

    True.
    Also true.
    Not sure what you can immediately deduce from this.
    You might want to reconsider that.
     
  6. May 18, 2007 #5
    Well, what I thought was the longer it takes to reach the ground, the more time it has to travel horizontally, therefore the more horizontal distance it travels. Skipping over, you get:
    The bigger the angle, the more horizontal distance it travels.
    So they are directly proportional.
     
  7. May 18, 2007 #6
    So what about the angle of 90 degrees? :-)
     
  8. May 18, 2007 #7
    Hmm. It does say it is a muzzle angle, so if the angle is 90, then you end up in shooting yourself and then you wouldn't care about the answer anyway. :)

    Yeah, I guess my non-formula logic doesn't apply for a 90 angle. Then the answer would be C.
     
  9. May 18, 2007 #8
    I just realized that if you throw it too high, then Vx will decrease.
    So yeah, all my posts don't make sense. Except for the one with the 90 degree angle. :)

    Could you say it is proportional to sin(2*angle), then?
     
    Last edited: May 18, 2007
  10. May 18, 2007 #9

    cristo

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    What makes you think that?
     
  11. May 18, 2007 #10
    Am I right?

    i too came to Husky's answer before posting the topic. But i had doubts whether it was right or wrong.

    let the angle be 'A'.
    Let the velocity be 'V'
    therefore
    horizontal component = V cos A
    vertical component = V sin A

    therefore
    horizontal displacement = V cos A * t(time of flight) units
    vertical displacement = V sin A * t units

    At maximum height,
    V sin A = 0. ----------------> l

    consider formula V = U-gt ----------------> ll


    therefore substituting l in ll

    0 = V sin A - gt
    t = (V sin A)/g

    total time of flight = 2t
    =2((V sin A)/g)

    range is nothing but
    the horizontal displacement V cos A * t

    so range = (2(V sin A)/g} * V cos A

    therefore range is directly proportional to Sin2A

    Am I right guys?
     
  12. May 18, 2007 #11
    dont just visit people
    please comment
    reply pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
     
  13. May 18, 2007 #12

    Astronuc

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    Staff: Mentor

    Last edited: May 18, 2007
  14. May 18, 2007 #13
    ya hooooooooooooooooooooo
    Thanks guys.
    Physics Forums Rok and you guys are cool
    the site u gave me is just beyond cool dude


    How much does wind affect the range of a projectile body?
    And how much does the shape and weight of the body affect the range?
     
    Last edited: May 18, 2007
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