Two stones are thrown vertically upwards with the same velocity of 49m/s. If they are thrown one after the other with 3 seconds in between, what is the height at which they collide?

Answer 1

1st: 49t-4.9t^2
2nd: 49(t-3)-4.9(t-3)^2

So, when are the heights equal?

49t-4.9t^2 = 49(t-3)-4.9(t-3)^2
t=6.5

Now just plug that in to find the height.