A ball is thrown vertically upward with an initial speed of 15 m/s. Then, 0.92 s later, a stone is thrown straight up ( from the same initial height as the ball) with an initial speed of 27 m/s. The acceleration of gravity is 9.8 m/s^2. How far above the release point will the ball and stone pass each other? Answer in units of m. So far, I have written down this equation 15( t + 0.92) - 4.9(t + 0.92)^2= 27t- 4.9t^2 And when I simplify everything, I get a t= .4593 Now, I don't know if that is even remotely correct, but if I am, I would like to know how to go about getting closer to the answer at hand. Thank you.