A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?

Answer 1

time that ball is in the air = t
time that stone is in the air =(t-2)
distance ball goes up = 20 t - 4.9 t^2
distance stone goes up = 24(t-2)-4.9 (t-2)^2
so
20 t -4.9 t^2 = 24 t - 48 - 4.9(t^2 - 4 t + 4)
20 t - 4.9 t^2 = 24 t - 48 - 4.9 t^2 + 19.6 t - 19.6
solve quadratic for t
use t in distance equation for either ball or stone to get height