I would like to know if a object is launched horizontally in the air at a lower speed and a higher speed, why does the higher speed keep it airborne longer. what are the forces and why does gravity suddenly take "longer"

And is this relevant to a pole dancer spinning around a pole? Why is it easier to stay high on the pole for longer if more momentum is used?


The time airborne is dependent only on the vertical component of velocity. It is described by the following equation when gravity is uniform:

$t_a = \frac{2v_{oy}}{g}$, where $v_{oy}$ is the vertical component of the projectile's velocity and g is the acceleration due to gravity.

If you launch an object horizontally, then it has no vertical component of velocity and its airborne time is only dependent on it's height above ground when it's launched:

$t_a = \sqrt{2h/g}$, where h is the height of the object when it is launched.

If the object has a larger component of horizontal velocity, it will travel farther during its time in the air, but as the above two equations show, the amount of time it spends in the air is not dependent on the value of its horizontal velocity.

This is only relevant to a pole dance if the pole dance is a physicist :)

Actually, what's probably happening on the pole is that a larger velocity around the pole contributes to a higher centripetal force for the dancer. Friction is proportional to the force between the dancer and the pole, so increased centripetal force leads to increased friction which means its harder to slide down.

  • $\begingroup$ All of the initial projectile motion analysis neglects air resistance, which is typical in a first course but not completely realistic with high velocities and non-trivial projectile sizes. $\endgroup$ May 21 '14 at 2:03
  • $\begingroup$ @dmckee air resistance wouldn't affect the time of flight, when starting from rest, right? The distance along horizontal would be effected, but not time of flight, assuming the objects are similar in shape. $\endgroup$
    – tpb261
    May 21 '14 at 6:01
  • $\begingroup$ @tpb261 It's been a while since I worked that out precisely. I believe that the correct answer depends on the functional form of the drag. I seem to recall that a $F \propto v^2$ drag rule (which is a pretty good approximation for, say, a baseball thrown at STP) means than the argument above remains good, but other rules cause variation in the falling time dependent on the horizontal velocity. Perhaps I'll find time to write a proper answer later on. $\endgroup$ May 21 '14 at 16:15
  • $\begingroup$ @dmckee " other rules cause variation in the falling time dependent on the horizontal velocity" - that's interesting. $\endgroup$
    – tpb261
    May 22 '14 at 3:33

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