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A diver running 1.2 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.1 s later. How high was the cliff and how far from its base did the diver hit the water?

  • Physics -

    vertical drop
    S = ut+(1/2)gt²
    = 0 + (1/2)(9.8)*(3.1)^2

    How far from base
    1.2m/s * 3.1s = 3.72m

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