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A stone is thrown from a 50m high cliff and lands 5 seconds later, 40m from the base of the cliff. At what speed and angle was the stone thrown?

  • Physics -

    The horizontal velocity component must be Vx = 40/5 = 8 m/s

    The initial vertical component Voy must satisfy this equation

    Voy*T - (g/2)T^2 = -50
    where T = 5 s, so

    Voy = 14.5 m/s

    Speed when thrown = sqrt(Voy^2 + Vx^2]
    = sqrt[(14.5)^2 + 8^
    = 16.6 m/s

    The ratio Voy/Vx is the tangent of the launch angle

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