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Thrown ball

  1. Apr 11, 2016 #1
    1. The problem statement, all variables and given/known data

    Ball A is thrown vertically by speed 40 m/s. After 1 second, Ball B is thrown vertically also by speed 47.5 m/s.
    What height from ground will they pass each other? ( g = 10m/s2 )

    2. Relevant equations

    s = vt + ½at2
    t = v/g
    v2 = v02 + 2as


    3. The attempt at a solution

    Time A to reach the max height is t = v/g = 40/10 = 4 s

    I don't understand. Because there is plus of 1 sec for ball B.
     
  2. jcsd
  3. Apr 11, 2016 #2

    SteamKing

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    What's not to understand?

    Both balls are thrown up in the same direction, although at different times and different speeds. The two balls are going to pass each other at some point, either on the way up or on the way down. At what height are the two balls the same distance above the ground?

    Hint: write a kinematics equation for each ball.
     
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