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March 27, 2017

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Multiple part question…..

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0m above a flat, horizontal beach. With what angle of impact does the stone land (in degrees).

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18 m/s. the cliff is 50 m above a flat, horizontally beach. how long after being released does the stone strike the beach below the cliff?

A student atands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above flat, horizontal beach. With what speed of impact does the stone land?

  • physics - ,

    The horizontal velocity component remains
    Vx = 18.0 m/s
    during the fall to the beach.

    The vertical component of velocity (measured positive upwards) is
    Vy = - gt

    The time t that it takes to reach the beach when falling from height H is given by solving
    H = (g/2) t^2

    Therefore t = sqrt (2H/g) = 3.19 s

    That answers your second question. Use that value of t to compute Vy at impact and from that and Vx get the angle of impact.

    The speed of impact is sqrt[Vx^2 + Vy^2]

  • Physics - ,

    Nutz

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