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Finding time a ball is in the air

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 15.44 m/s when the hand is 1.61 m above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

    2. Relevant equations

    I need help starting this problem

    3. The attempt at a solution
     
  2. jcsd
  3. May 13, 2009 #2

    LowlyPion

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    Homework Helper

    See:
    https://www.physicsforums.com/showpost.php?p=905663&postcount=2
     
  4. May 13, 2009 #3
    [tex]v_{f}=v_{o}+at[/tex]
    [tex]0m/s = 15.44 m/s +(-9.81m/s^2)*t[/tex]
    [tex]t = 1.574 s[/tex]

    [tex]v_{f}^2=v_{o}^2+2a*x[/tex]
    [tex]0 = 15.44^2m/s + 2 *(-9.81m/s^2) * x[/tex]
    [tex]x = 12.15 m[/tex]
    This is how high the ball travels from the hand.

    -------------When the ball is returning to earth-----------
    [tex]x = x_{0} +v_{0}t+\frac{1}{2}at^{2}[/tex]
    [tex]x = 12.15m + 1.61m = 13.76m[/tex]
    [tex] 13.76m = 0m + 0m/s*t + \frac{1}{2} * 9.81m/s^2*t^2[/tex]
    [tex] 27.52 = 9.81m/s^2*t^2[/tex]
    [tex]2.805 = t^2[/tex]

    [tex]t = 1.675 s[/tex]

    [tex]t_{total} = t_{ascent} +t_{decend}[/tex]

    [tex]t_{total} = 1.975s + 1.574 s[/tex]

    [tex]t_{total} = 3.549s[/tex]
     
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