physics
posted by morgan on .
Two students are on a balcony 24.4 m above the street. One student throws a ball, b1, vertically downward at 20.3 m/s. At the same instant, the other student throws a ball, b2, vertically upward at the same speed. The second ball just misses the balcony on the way down.
(a) What is the difference in time the balls spend in the air?
(b) What is the velocity of each ball as it strikes the ground?
velocity for b1
velocity for b2
(c) How far apart are the balls 0.510 s after they are thrown?

height of b1:
h1 = 24.4  20.3t  4.9t^2
h1=0 when t = 0.973 sec
h2 = 24.4 + 20.3t  4.9t^2
h2=0 when t = 5.116 sec
v1 = 20.3  9.8t
v2 = 20.3  9.8t
plug in the values for t to get the final velocities and heights at 0.510 sec