A stone is thrown vertically upward at a speed of 17.4 m/s at time t=0. A second stone is thrown upward with the same speed 0.56 seconds later. At what time are the two stones at the same height?

At what height do the two stones pass each other?

What is the upward speed of the second stone as they pass each other?

You need to write out the equations of motion for the two stones, using the basic formula:


y = y_0 + v_0t + \frac 1 2 a t^2


You're given the initial velocity v_0, and acceleratoin a is equal to -g . So the only trick here is to use t-0.56 in place of t for the equation of the second stone. Set the two equations equal, and solve for t. Can you take it from here?