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Projectile Motion balloon question

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 23.5m/s A 1.0-kg stone is thrown from the basket with an initial velocity of 14.8m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. The person in the basket sees the stone hit the ground 6.20s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 23.5m/s.

    How high was the balloon when the rock was thrown out?

    2. Relevant equations
    ΔY=V1y*t + 1/2*a*t^2

    3. The attempt at a solution
    At first I thought since the stone is being thrown off the balloon which is going at a constant speed of 23.5m/s, then the stone's initial Y velocity must also be 23.5m/s, and since its being thrown perpendicular to the balloon's path at 14.8m/s, then that must be its initial X velocity.

    I figure that the balloon's height when the stone was thrown out, is the same as the stone's initial position. And the stone's final position is 0 at t=6.20 s since its on the ground.

    So I used the above equation
    ΔY=V1y*t + 1/2*a*t^2

    0-y1=V1y*t - 1/2*g*t^2

    y1= -[(23.5m/s)(6.20s)-(4.9m/s^2)(6.20s)^2
    y1=47.356m

    but this isn't right and I don't know why
     
  2. jcsd
  3. Mar 1, 2015 #2

    haruspex

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    It says 'perpendicular to the path of the descending balloon'. What direction does that make it?
     
  4. Mar 1, 2015 #3
    oh wow ok. The initial Y velocity is negative. I got 334m
     
  5. Mar 1, 2015 #4
    Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest in the basket.

    I'm having trouble figuring out the above part. I know the X velocity is constant, but i dont know how to get the Y velocity. I tried using the equation V^2=V1^2 + 2aΔY

    V^2=(-23.5m/s)^2 -2*(9.8m/s^2)*188m

    and i know this is wrong so idk
     
  6. Mar 1, 2015 #5

    haruspex

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    yes, that looks about right.
     
  7. Mar 1, 2015 #6

    haruspex

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    What was the y velocity relative to the person in the basket when the rock was thrown? What was the y acceleration relative to the basket?
     
  8. Mar 1, 2015 #7
    would it be

    V^2=-2*(9.8m/s^2)*(-188m)
    V=60.7 but it would be negative since the velocity is in the negative direction?
     
  9. Mar 1, 2015 #8

    haruspex

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    Where are you getting the 188m from?
     
  10. Mar 1, 2015 #9
    That's what I calculated for the Y displacement of the stone when it reached the ground, relative to the balloon.
     
  11. Mar 1, 2015 #10

    haruspex

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    Ok. Yes, that's the right answer. Your method produces an ambiguity in the sign. You can avoid that (and the need to take a square root) by using the time and relative acceleration.
     
  12. Mar 1, 2015 #11
    Thank you, and your right that would've been a much easier way. I tend to over complicate things more than I should.
     
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