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? asked in Science & MathematicsPhysics · 2 years ago

Physics help please?

A diver jumps up off a diving board with an initial velocity of 3.75m/s. What is their velocity when they reach the water 4.0m below? (assume perfect physics)

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  • oldschool
    Lv 7
    2 years ago
    Favorite Answer

    At launch the diver's energy = mgh+½mV² = m(gh+V²/2)

    m(gh+V²/2) = = m(9.8*4+3.75²/2) = 46.23*m J

    At h = 0

    ½mV² = 46.23*m J

    V² = 2*46.23 = 9.6²

    V = ~ 9.6m/s

    The launch angle does not matter because V² = Vx²+Vy²

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