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A cement block accidentally falls from rest from the ledge of a 51.7-m-high building. When the block is 14.9 m above the ground, a man, 2.05 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?

  • physics -

    free fall from 51.7 to 14.9 meters - how fast is it going?
    easy way is energy
    (1/2) m v^2 = m g (51.7-14.9)
    v^2 = 2 * 9.81 * 36.8
    v = 26.7 m/s initial speed down at 14.9 height
    how long to fall another (14.9 -2.05)= 12.85 meters?
    increase in Ke = m g h
    (1/2) m v^2 = (1/2)m(26.7)^2 + m g (12.85)

    v^2 = 2* 26.7^2 + 2*9.81(12.85)
    v^2 = 1677
    v = 41 m/s when it hopefully misses
    average speed between sighting and missing = .5(41 + 26.7) = 33.85 m/s
    time to fall 12.85 m = 12.85/33.85 = .38 seconds (forget it)

  • physics -

    Your formula is wrong, i tried it twice already.

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