# Question #13df4

Mar 5, 2016

No, these are two very different effects. Sometimes, like in free-fall in the atmosphere, they come to an equilibrium.

#### Explanation:

Very often, in high-school level and early university or college physics courses, the effect of air resistance is neglected in projectile motion problems. This makes the problems much easier to solve. The two are quite unrelated. The force due to gravity is given by

${F}_{g} = m \cdot g$

Where $m$ is the mass of the object and $g$ is the acceleration due to gravity ($g = 9.8 m / {s}^{2}$ here on Earth) Air resistance, or drag, can be calculated from the following formula:

${F}_{\mathrm{dr} a g} = - b \cdot v$

Where the negative sign implies that the force always opposes motion, $b$ is a drag coefficient for the object which depends on it's shape and size and $v$ is the velocity. We can see that the drag depends on the velocity of the object.

Often, air resistance can be ignored. For relatively dense objects moving at reasonable speeds over short distances and times, the effect of air resistance is quite small. A good example of this would be a basketball thrown by a player toward the basket. The result of calculating the arc without air resistance is actually quite good.

Sometimes, it doesn't make sense to ignore air resistance, like in the case of free-fall of an object from a great height. For instance, a parachute jumper falling back to earth. Air resistance increases with increasing velocity, therefore an equilibrium obtained where the drag due to air resistance equals the force due to gravity at some velocity. This is usually called the object's terminal velocity. The following page does a good job of describing this situation:
http://www.physicsclassroom.com/class/newtlaws/Lesson-3/Free-Fall-and-Air-Resistance