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A cement block accidentally falls from rest from the ledge of a 70.7 -m-high building. When the block is 18.0 m above the ground, a man, 2.10 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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let the time taken to reach 18 m above the ground (52.7 m from the initial posiiton be x.so,

H=u*t + 0.5gt^2

52.7 = 0*x + 0.5*9.81*x^2

or x=3.277

let the time taken to reach 2.1 m above the ground (68.6 m from the initial posiiton be y).so,

H=u*t + 0.5gt^2

68.6 = 0*y + 0.5*9.81*y^2

or x=3.73

so time left for him = 3.73-3.277

=0.453 s
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