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5 August, 03:56

Two people are standing on the edge of a building that is 42 meters high. One person throws a tennis ball straight downward at a speed of 16 m/s. At the same exact time, the other person throws a tennis ball straight upward at 16 m/s. How long after the first tennis ball lands will the second tennis ball arrive at the ground?

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  1. 5 August, 04:13
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    Answer : 1.67 sec

    Explanation:

    Time calculation for person 1

    Height of the building = 40meters

    Initial velocity = 16m/s

    Final velocity = 0m/s

    Accelaeration = 9.8m/s2

    Using first equation of motion

    V=u+at

    0 = 16 + 9.8t

    -16/-9.8 = t

    Time = 1.63 seconds

    Time calculation for person 2

    While the ball is going up

    Height of the building = 40meters

    Initial velocity = 16m/s

    Final velocity = 0m/s

    Accelaeration = - 9.8m/s2

    Using second equation of motion

    v2 - u2 = 2as

    0-256 = - 19.6s

    S = - 256/-19.6 = 13.06m

    Total distance. = 13.06 + 40 = 53.06 meters

    Using third equation of motion

    S = ut + ½ * a * t*t

    S = 0 + ½ * 9.8*t*t

    T2 = 53.06/4.9

    T2 = 10.83

    T =  10.83 = 3.3 sec

    Diffenrece of time = 3.3 - 1.63 = 1.67 sec
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