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Hot Air Balloon Free Fall Question

  1. Sep 5, 2004 #1
    A hot-air balloonist, rising vertically with a constant velocity of magnitude v=5.00m/s , releases a sandbag at an instant when the balloon is a height h=40.0m above the ground . After it is released, the sandbag is in free fall.

    1.) Compute the position of the sandbag at a time 0.165s after its release. Take the free fall acceleration to be g=9.80 m/s^2. answer ___ m.

    how would i find that? i dont know where to start...

    i tryed to find the highest time that it takes, 5.00/9.8 = 25/49

    im just doing random things for hours, can someone point me to the correct way?
     
  2. jcsd
  3. Sep 5, 2004 #2
    For one dimensional motion with constant acceleration (a) the following formula gives you the position (x(t)) of the object as a function of time (t):

    [tex]x(t)=x(0)+v(0)t+\frac{1}{2}at^2 [/tex]

    Where v(0) is the initial velocity of the object. Be careful with the signs of the paramters!
     
  4. Sep 5, 2004 #3
    so...
    x(t)= 40m + 5.00 + 1/2(9.80)(0.165)^2 = 45.33m <-- am i doing it correctly?
     
  5. Sep 5, 2004 #4
    You forgot to multiply the initial velocity with time. And also, you chose your inital velocity positive meaning upwards, but the acceleration has another direction, downwards. So you have to use a=-9,8 m/s^2.

    [The formula comes from Newtons second law F=ma, with F the gravitational force mg, so mg=ma -> g=a=v'=x'' where ' denotes a derivative with respect to time. When you integrate the constant g two times with respect to time (using the appropriate boundry conditions (initial position/velocity)) you get the position as a function of time, the formula I gave you.]
     
  6. Sep 7, 2004 #5
    thanks, that worked. i have another question.

    Compute the velocity of the sandbag at a time 0.165s after its release. Take the free fall acceleration to be g=9.80m/s^2 .

    i just need to know the formula i should use.

    i tried this formula: v(x) = sqrt(v(0)^2+2ax)) which doesnt work
     
  7. Sep 7, 2004 #6
    velocity is the derivative of the poition to time:

    [tex]\frac{d}{dt}[ x(t)=x(0)+v(0)t+\frac{1}{2}at^2 ] [/tex]
    [tex]v(t)=\frac{dx(t)}{dt} = v(0) + at [/tex]

    Again, wach out with the signs...
     
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