A rock is thrown vertically upward with a speed of 12.0m/s.Exactly 1.00s later a ball thrown up vertically along the same path with a speed of 18.0m/s.

Question

a. At what time will the ball collide with the rock?

Answer 1

for the rock ... h = -4.9 t^2 + 12.0 t

for the ball ... h = -4.9 (t - 1)^2 + 18.0 (t - 1)

set the two heights equal , and solve for t