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Physics motion webassign

  1. Sep 9, 2010 #1
    1. The problem statement, all variables and given/known data
    A tennis ball is thrown straight up with an initial speed of 25.0 m/s. It is caught at the same distance above the ground.

    (a) How high does the ball rise?
    _________m

    (b) How long does the ball remain in the air? (Hint: The time it takes the ball to rise equals the time it takes to fall.)
    _________s


    2. Relevant equations



    3. The attempt at a solution
    i have no idea how to do this. please help!
     
  2. jcsd
  3. Nov 14, 2010 #2
    here's your first equation:
    T-time Vf-final velocity
    Vi-initial velocity A-acceleration
    T=(Vf-Vi)/A
    the object is going up, so all Velocity is up (no horizontal motion)
    vertical acceleration is always -9.8 (due to gravity)
    Vi=25m/s, and Vf is 0 at the top (where max vertical distance is reached) so,
    T=(0-25)/-9.8 OR T=25/9.8 <(T for max height)
    simple multiply by two, and you have your answer for number 3.

    your second equation would be:
    Dy-change in y distance (vertical), T, Vi, A, and angle (90)
    Dy=Vi(sin(90))t+1/2at^2 so,
    Dy=25(1)t-4.9t^2 OR Dy=25(25/9.8)-4.9(25/9.8)^2

    therefore, T~5.102 <(total T) and Dy~31.888m
    where ~ means approximately equal to
     
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