7. a. distance travelled by ball 1 = 0 + 9.8 t2 /2 = 4.9t2 distance travelled by ball 2 = 20t - 4.9t2 distance by ball 1 + distance by ball 2 = 67m (4.9tz) + (20t - 4.9t 2) = 67m 20t = 67m t = 3.35s h = 20t - 4.9t 2 = 20(3.35) -4.9(3.35) = 67 - 54.99 = 12.0 m, therefore the collide at 12.0m above the ground