# physics

Two students are on a balcony 18.3 m above the street. One student throws a ball (ball 1) vertically downward at 12.2 m/s; at the same instant, the other student throws a ball (ball 2) vertically upward at the same speed. The second ball just misses the balcony on the way down.

(a) What is the difference in the two ball's time in the air?

(b) What is the velocity of each ball as it strikes the ground?
ball 1

ball 2

(c) How far apart are the balls 0.800 s after they are thrown?

cannot seem to figure out how to find part a) i cannot get the second balls time.

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1. a)s=vt so t=s/v t=18.3m/12.2ms=1.5s the same is for the other so t1=1.5s & t2=1.5s so t1-t2= 1.5s-1.5s=0

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posted by John

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