physics

a stone is from the top of a tower 300m high and at the same time another stone is projected vertically upward with initial velocity 75m/s. calculate when and where the two stone meet

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  1. d1 + d2 = 300m,
    0.5gt^2 + (75t + 0.5gt^2) = 300m,

    Since g in the 1st term is positive and
    g in the 3rd is negative, they cancel
    each other and the Eq becomes:

    75t = 300,
    t = 4 seconds.

    d1 = 0.5*9.8*4^2 = 78.4m. below top
    of tower = 300 - 78.4 = 222m above
    ground.

    d2 = 75*4 - 0.5*9.8*4^2,
    d2 = 300 - 78.4 = 222m above ground.

    Therefore, the 2 stones meet 4s after
    release at a distance of 222m above
    ground.

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