A cement block accidentally falls from rest from the ledge ofa 53.5 m-high build

Question

A cement block accidentally falls from rest from the ledge ofa 53.5 m-high building. When the block is 13.1 m above theground, a man, 1.96 m tall, looks up and notices that the block isdirectly above him. How much time, at most, does the man haveto get out of the way?   (Units are in seconds)     A cement block accidentally falls from rest from the ledge ofa 53.5 m-high building. When the block is 13.1 m above theground, a man, 1.96 m tall, looks up and notices that the block isdirectly above him. How much time, at most, does the man haveto get out of the way?   (Units are in seconds)    

Explanation / Answer

let time of travel from height 53.5m to 13.1m ist1 then (53.5m - 13.1m) = (1/2)(9.8m/s2)t12 then t1 =[(40.4m)(2)/(9.8m/s2)]           =2.87s let time of travel from height 53.5m to 1.96m ist2 then (53.5m - 1.96m) = (1/2)(9.8m/s2)t22 then t2 =[(51.54m)(2)/(9.8m/s2)]           =3.24s So time he had is 3.24s - 2.87s = 0.37s           =2.87s then t1 =[(40.4m)(2)/(9.8m/s2)]           =2.87s let time of travel from height 53.5m to 1.96m ist2 then (53.5m - 1.96m) = (1/2)(9.8m/s2)t22 then t2 =[(51.54m)(2)/(9.8m/s2)]           =3.24s So time he had is 3.24s - 2.87s = 0.37s           =2.87s let time of travel from height 53.5m to 1.96m ist2 then (53.5m - 1.96m) = (1/2)(9.8m/s2)t22 then t2 =[(51.54m)(2)/(9.8m/s2)]           =3.24s So time he had is 3.24s - 2.87s = 0.37s           =2.87s then t2 =[(51.54m)(2)/(9.8m/s2)]           =3.24s So time he had is 3.24s - 2.87s = 0.37s           =2.87s

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