A skateboarder rolls off a ledge that is 1.12 m high, and lands 1.48 m from the base of the edge. How much time was he in the air?

The rider's horizontal motion, and how much ground he covers before he hits it, have nothing to do with how long he takes to hit the ground.  The problem is simply:  "How long does it take an object to fall 1.12 m from rest ?"

This seems like a good time to use this formula:

Distance fallen from rest = (1/2) (acceleration) (time)²

The problem doesn't tell us what planet the skateboarder is exercising  on. I'm going to assume it's on Earth, where the acceleration of gravity is 9.8 m/s².  And now, here's the solution to the problem I just invented:

1.12 m = (1/2) (9.8 m/s²) (time)²

Time²  = (1.12 m) / (9.8 m/s²)

Time²  = 0.1143 sec²

Time = √(0.1143 sec² )

Time = 0.34 second

newton's first law of motion