u(y) = 0
s(y) = ut + 1/2 at^2
s = 24t + 4.8t^2
s is the vertical displacement from its initial location.
If you just sub in s=50 here, that will find the time for it to reach 100m.
To find the time it takes to reach the ground, you need to sub in s = -50
I would do the rest but I take that's...