A man is standing on the top of the building 44.1m high throws a stone upward with a certain initial speed. After 2.0s another stone is dropped from the same point. What must be the initial speed of the first stone so that they will both reach the ground at the same time?

You have to know your formulae;

s = ut + \frac12 at^2

Find the time t for the second stone to fall down, given the displacement s, 44.1, the acceleration due to gravity, a, 9.81 m/s^2 and initial velocity, u, 0 m/s

Now, fit that time added with 2 seconds into your formula, putting 'a' as negative and solve for u, the initial speed.