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

this is an example of "committed" time.

committed time, as contracted time, takes need over essential and available time since it is seen as profitable work. it alludes to the time allotted to keep up a home and family. at the point when a man is commuting using committed time this individual may feel that the drive is an obligation to family, for example, strolling youngsters to class or driving a mate to work. contracted and committed time clients may feel that their commute could really compare to the commute of important or leisure time clients on the grounds that their commute is profitable work. in this manner, they might be more disposed to pick a mechanized method of movement.