A cement block accidentally falls from rest from the ledge of a 53.4-m-high building. When the block is 19.4 m above the ground, a man, 2.00

Question

A cement block accidentally falls from rest from the ledge of a 53.4-m-high building. When the block is 19.4 m above the ground, a man, 2.00 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way

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Thu Cúc 6 months 2021-07-17T23:27:24+00:00 1 Answers 1 views 0

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    2021-07-17T23:28:52+00:00

    Answer:

    The time required by the man to get out of the way is 0.6 s.

    Explanation:

    height of building, H = 53.4 m

    height of block, h = 19.4 m

    height of man, h’ = 2 m

    Let the velocity of the block at 19.4 m is v.  

    use third equation of motion

    v^2 = u^2 + 2 gh\\\\v^2 = 0 + 2 \times 9.8 \times (53.4 - 19.4)\\\\v = 25.8 m/s

    Now let the time is t.

    Use second equation of motion

    h = u t + 0.5 gt^2\\\\19.4 - 2 = 25.8 t + 4.9 t^2\\\\4.9 t^2 + 25.8 t - 17.4= 0 \\\\t = \frac{-25.8\pm\sqrt{665.64 + 341.04}}{9.8}\\\\t = \frac{-25.8\pm31.7}{9.8}\\\\t = 0.6 s, - 5.9 s

    Time cannot be negative so time t = 0.6 s.

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