Consider the two linear expressions.
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Answer is 3 seconds
When the bullet reaches the ground, ground being x in graph (and here its s which is = 0)
s = -16t^2 + 48t
s = 0, solve for t
0 = -16t^2 + 48t
0 = t ( -16t + 48)
0 = 16t ( - t + 3)
now you have two equation
0 = 16t and 0 = -t +3 ( you can look at the graph line touches x twice)
0 = 16 t
0 = t ( you know its false, because time = 0)
You are left with
0 = -t + 3
t = 3
It takes 3 seconds for the bullet to return to the ground.
// Hope this helps.
