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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?

Added Thu, 30 Jul '15

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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

Added Thu, 30 Jul '15

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