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A diver running 1.8m/s dives out horizontally from the edge of a vertical cliff and 3.0s later reaches the water below. How high was the cliff, and how far from its base did the diver hit the water?

  • physic - ,

    V initial for x axis: 1.8
    V initial for y axis: 0.0
    Acceleration on x axis: 0
    Acceleration on y axis:-g=-9.81
    Δ time: 3
    Y final:0
    X initial: 0
    looking for: Y initial and X final

    eqt 2-11b: Δx=Vi(Δtime)+(a*Δt^2)/2

    for x axis:
    Δx=Vi(Δt)+ zero (canceled out)
    = 1.8(3)=5.4
    5.4 m far from base

    for y axis:
    Δy= zero (canceled)+(a*Δt^2)/2
    -44.1m is your height of cliff

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