Posted by Chris on .
Two students are on a balcony 24.9 m above the street. One student throws a ball, b1, vertically downward at 19.6 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= m/s
velocity for b2 = m/s
(c) How far apart are the balls 0.440 s after they are thrown?
hf=ho+Vi*t-1/2 g t^2
-24.9 = Vi*t-1/2 g t^2 use the quadratic eqution.
solve for t, given Vi is +, then -
then subtract the difference.