A stone is thrown vertically upward from the roof of a building with velocity of

Question

A stone is thrown vertically upward from the roof of a building with velocity of 29.4m/s. Call t=0 the moment it is thrown. Another stone is dropped 4s later, it has zero initial velocity and falls down. At what time does the first stone pass the second stone? At what height?

Explanation / Answer

vi=voy=29.4 m/s--->max ht t=voy/g=29.4/9.8=3 s---> max ht=voy^2/(2*9.8)=44.1 m or 29.4*3+1/2*-9.8*3^2=44.1 m--->same ans. it takes 6 s to fall to initial pt at voy when voy 2nd stone=voy 1st sone--->at 2.4672 s---> 1.29 m from roof---> 29.826 m/s

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