Ball and stone thrown upward at different times—where do they meet?

[AraProdieur]

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.

 

[learningphysics]

EDIT: oops sorry... for some reason I thought the second ball was being thrown down from above. Thanks Astronuc.

 

[Astronuc]

Your solution is correct for finding time, t, which according to your approach is the time after release of the stone. The ball is leading by 0.92 s.

So one has set the height equal and then one solves for the time.

Then pick one of the height equations and solve for the height as a function of the t.

Since one has a quadratic, make sure to pick the correct root.