University Physics

A rock is tossed straight up with an initial speed of 20 m/s. When it returns, it fall into a hole that is 8.0 m deep. What is the velocity of the rock (in m/s) when it hits the bottom of the hole? Use -9.80 m/s2 for the local acceleration due to gravity and assume no significant air resistance. Assume that the y axis is positive in the up direction

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  1. -8 = 0 + 20 t - 4.9 t^2

    4.9 t^2 -20 t -8 = 0
    solve quadratic for t

    v = 20 - 9.8 t

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  2. I am sorry but I am still confused...

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  3. more details:

    h = original height + Vi t - (1/2)(9.8) t^2

    -8 = 0 + 20 t - 4.9 t^2

    4.9 t^2 -20 t -8 = 0
    solve quadratic for t

    velocity = original velocity - 9.8 t
    v = 20 - 9.8 t

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    2. 👎 0

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