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Projectile motion question

  1. Aug 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A hang gilder runs off a cliff, with a speed of 3 m/s. What is its overall velocity after 5 seconds?

    Some constants like gravity are rounded because this is an MCAT problem.

    2. Relevant equations
    v=v(initial) + at
    a^2 + b^2 = c^2

    3. The attempt at a solution

    vertical velocity:
    v=v(initial) + at
    0=v(initial) + at
    v(initial) = -at
    = (-10 m/s)(5s)

    What i find strange is that in the solution manuel for this problem, 0.5 seconds is used for time, giving the overall vertical velocity -5 m/s.

    Then you find the overall velocity with pythagorean's theorum, but what is strange is instead of using 3 m/s for the horizontal velocity, 9 m/s is used instead! :
    v= (9^2 + -5^2)^(1/2)
    Making it just a tad over 10.

    Why is 0.5 seconds used instead of 5 seconds?

    And why is 9 used instead of 3??

    Thank you!
     
  2. jcsd
  3. Aug 4, 2009 #2
    You manipulated the equation wrong, and your units are wrong as well. The units for acceleration are [tex]\frac{meters}{second^2}[/tex]

    I have absolutely no idea how your book/coursework or whatever reached its answers.
    A substitution of [tex]t=5 sec[/tex] in the [tex]v(t)[/tex] function was required. No clue on why they chose to substitute with [tex]t=0.5 sec[/tex]
    Same goes for the magnitude of the velocity vector, the horizontal velocity remained constant at [tex]3 \tfrac{m}{s}[/tex] so that should have been substituted into the equation, and not the nonsensical [tex]9 \tfrac{m}{s}[/tex]

    You were on the right track, a few minor errors aside, so I don't have any qualms about posting the full solution.

    In order to find his vertical velocity after 5 seconds:
    [tex]v(t)=v_i+at[/tex]

    Now we substitute for the question's known variables.
    [tex]v(t)=0-gt[/tex]

    Calculating for [tex]t=5[/tex] yields:
    [tex]v(5)=-g*5=-50 \tfrac{m}{s}[/tex]

    As you've correctly pointed out, his horizontal velocity remains constant as there are no forces acting on him in that direction.

    Some quick vector math gives us the magnitude of the velocity vector:
    [tex]|\vec v|=\sqrt{|\vec v_x|^2+|\vec v_y|^2}[/tex]
    Substituting:
    [tex]|\vec v|=\sqrt{3^2+50^2}=\sqrt{2509}\approx 50.0899 \tfrac{m}{s}[/tex]
     
  4. Aug 4, 2009 #3

    ideasrule

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    Homework Helper

    This is a hang glider, and it's supposed to be in free-fall? This is probably the worst physics problem I've ever seen!
     
  5. Aug 4, 2009 #4
    Doesn't seem like a very nice problem, because hang-gliders are assumed to actually fly and glide, instead of "running and plummeting".
     
  6. Aug 4, 2009 #5
    Maybe the guy who made the glider was the same guy who made the problem.
     
  7. Aug 4, 2009 #6

    kuruman

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    Homework Helper
    Gold Member

    Then he should jump off the cliff without the glider. This would make the problem more consistent with projectile motion. My apologies about the sarcasm, but this is indeed a terrible problem.
     
  8. Aug 4, 2009 #7
    Maybe his problem is that his hang-glider doesn't quite glide?

    Anyway, a model including simple air resistance (Ignoring lift and other pressure effects), could do nicely here as well, but as you're not given any constants for the drag force, that's out of the question as well. Where exactly are you getting these questions? You should very seriously consider using a different book/tutor/website.
     
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