Question

A stone is thrown vertically up from the top of the tower 40m high with the velocity of 20m/s. 3 seconds later another stone is thrown vertically up from the ground with the velocity of 30m/s.Calculate when and where the two stones will meet from the foot of the tower.

Answer 1

stone 1: height is
40 + 20t - 9.8t^2

stone 2, 3 seconds later:
30(t-3) - 9.8t^2

they meet (or pass closely) when the heights are equal:

40 + 20t - 9.8t^2 = 30(t-3) - 9.8(t-3)^2

So, just solve that for t, and then plug that value into either equation.