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Initial velocity of hot air ballon, and baseball

  1. Nov 4, 2007 #1
    Question: A passenger in a hot-air balloon throws a ball with an initial unknown velocity. The ball accelerates at 9.5m/s2 (D) for 2.0s. At the 2.0-s point, its instantaneous velocity is 24m/s [45 degrees below the horizontal]. Determine the ball's initial velocity.

    I don't quite understand what the question is asking for. I 'think' what the question wants is how fast the hot air balloon is moving, which equals the balls initial velocity...is this right?

    The diagram below shows the movement plotted on a graph. So to find the ball's initial velocity, really all I'm looking for is the velocity of the X component of the vector, right?



    Vx = cos 45degree (V)

    =17m/s2 ???
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 4, 2007 #2


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    Staff: Mentor

    Well, no. The question is worded strangely, and the 9.5 m/s2 is interesting because normally the acceleration of gravity is 9.81 m/s2 (at sealevel), so does this infer air resistance perhaps or something about the elevation.

    One has to determine the direction of the acceleration, which if due to gravity is straight down if there is no thrust on the ball.

    Now there are downward (vertical) and horizontal components to velocity.

    Without any acceleration in the horizontal direction, the horizontal component of velocity is constant, and would be given by vx = v cos (45°). However, if the acceleration of 9.5 m/s2 implies air resistance, then there could be a deceleration due to air resistance of 0.3 m/s2.

    With respect to vertical (downward) motion, one can determine the velocity after 2 sec for a downward acceleration of 9.5 m/s2 and compare to the downward velocity component v sin (45°). The difference would be the initial velocity in the vertical direction.
  4. Nov 4, 2007 #3
    thankx for the quik reply......

    but ya, opps........made a typo.........no air resistance, just 9.8m/s2.

    I understand your breakdown of velocity in the X component, and velocity of Y component.

    but I still don't understand what the question is asking.

    maybe given the answer from the back of the back, you could help me understand what the question is asking for.

    Vf= 25 m/s [45 degrees D]
    a = 9.8 m/s2 [D]
    t= 2.0s

    Let up and right be positive

    Therefore, we can calculate the value of Vx as follows"

    Vx = v cos45degrees
    = 25 m/s cos45degrees
    = 16.97 m/s

    Since there is no horizontal acceleration, therefore, the initial horizontal velocity equals the final horizontal velocity, equals 17 m/s.

    to calculate the vertical component of the initial velocity:

    ay = Vyf - Vyi / t

    therefore Vyi = Vyf - ay(t)


    Vyf = Vsin(45degrees)
    = 24m/s(sin)(45)
    = 16.97m/s

    therefore Vyi = 16.97m/s - (9.5 m/s2)(2.0s)
    = 2.6 m/s

    Therefore, to calculate the value and direction of the initial velocity:

    Vi2 = Vix2 + Viy2

    = (17m/s)2 + (2.6 m/s)2

    therefore Vi =17 m/s

    To calculate the angle of the ball's velocity"

    tan(theta) = Vyi / Vxi

    = 2.6 / 17

    therefore theta = tan-1 2.6 / 17
    = 8.7 degrees
  5. Nov 4, 2007 #4


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    Staff: Mentor

    Hold on! How does one obtain Vi = 17 m/s if one is adding (2.6 m/s)2 to (17m/s)2 then taking the square root?

    Please review one's calculations.
  6. Nov 4, 2007 #5

    in my book, it teaches me to plot on a graph, Vxi and Vyi...thus giving Velocity initial as the connecting line, making a triangle.

    then using Pythagoreans theory, Vi2 = Vxi2 + Vyi2
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