Posted by Kelly on Friday, May 13, 2011 at 12:08pm.
Two students are on a balcony 19.6 m above the street. One student throws a ball (ball 1) vertically downward at 15.8 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?
(c) How far apart are the balls 0.500 s after they are thrown?
Please help!

Physics  bobpursley, Friday, May 13, 2011 at 12:18pm
a) find the time in air:
hf=hi+vi*t1/2 g t^2 hf=0, hi19.6, g=9.8m/s^2 solve for time, one vi is +, the other .
b) Vf=Vi*tg*t you know t.
c) when the downward ball hits, you know that time t. Find where the second ball is at that same time.
h=hi+Vi*t1/2 g t^2 
Physics  bobpursley, Friday, May 13, 2011 at 12:18pm
c) oops, find hi for each ball at time t=.5sec, then subtract them.