Here is another question.I try to get the answer but ..... 1.A ball is trown vertically upward with an initial speed of 15m/s.then,2.2 s later ,A stone is thrown straight up (from the same initial height as the ball)with an initial speed of 21.3m/s. the acceleration of g=9.8m/s^2. How far above the release point will the ball and stone pass each other? Here is what I did. I use the same time for both (2.2second). the equation is Y=Y(inital) + v(inital)t + 1/2 gt^2 After i got both distance I subtracted. which is gave me the didtance b/n them.