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A ball is thrown vertically upward with a velocities of 18 m/s. Two seconds later another ball is thrown upwards with a velocity of 13.5 m/s. At what position above the ground will they meet?

  • math -

    you want to find where

    18t - 4.9t^2 = 13.5(t-2) - 4.9(t-2)^2
    t = 3.086 seconds
    height: 8.883 m

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