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dizliz24

  • 4 years ago

I am a little confused on how to solve this, could someone help me out? A cement block accidentally falls from rest from the ledge of a 73.2-m-high building. When the block is 17.0 m above the ground, a man, 1.70 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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  1. Annand
    • 4 years ago
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    Here initial vel, u=0 as it falls from rest accl, a=-g=-9.8 So \[S=u*t+0.5*a*t^2\] becomes \[S=-0.5*9.8*t^2\]

  2. Annand
    • 4 years ago
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    Time to reach 17m, t1: Here displacement, S=17-73.2 t=t1 Substitute in the above eqn and find t1...

  3. Annand
    • 4 years ago
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    Time to reach 1.7m, t2: S=1.7-73.2 ; t=t2 find t2....

  4. Annand
    • 4 years ago
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    Time of travel from 17m to 1.7 m= t2-t1 He must evade before this time to save himself...

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