Free Fall Equations

AnnaWhen people hear “free fall,” they often picture anything dropping. But in physics, it has a more specific meaning, doesn’t it?

MarcoYes. Free fall means gravity is the only force acting on the object. In many school problems, we ignore air resistance. If air resistance matters, another force is affecting the motion, so it is not ideal free fall.

AnnaAnd near Earth’s surface, gravity gives objects an acceleration of about nine point eight meters per second squared downward. That’s acceleration, not speed.

MarcoExactly. It means the downward velocity changes by about nine point eight meters per second every second. So an object released from rest reaches a downward speed of about nine point eight meters per second after one second, if we ignore air resistance.

AnnaDoes that acceleration depend on how heavy the object is?

MarcoNot near Earth, as long as air resistance is negligible. A heavy object and a light one, dropped from the same height at the same time, have the same motion.

AnnaWe treat gravity as constant for ordinary falling problems because the height change is small compared with Earth’s radius. But the value can vary a little with location and altitude.

MarcoRight. And a more precise problem may give you a particular value rather than asking you to use nine point eight.

AnnaLet’s talk equations. The three standard ones are final velocity equals initial velocity plus acceleration times time; displacement equals initial velocity times time plus one half acceleration times time squared; and final velocity squared equals initial velocity squared plus two times acceleration times displacement.

MarcoThe letters matter, too. U is initial velocity, V is final velocity, A is acceleration, T is time, and S is displacement in the direction we choose as positive. How do you keep the signs straight?

AnnaChoose a positive direction first and stick with it. If downward is positive, acceleration is positive nine point eight meters per second squared. If upward is positive, acceleration is negative nine point eight. The displacement and velocities need signs that match the same choice.

MarcoAnd the units should fit: meters for displacement, seconds for time, meters per second for velocity, and meters per second squared for acceleration. A dropped object starts from rest, so its initial velocity is zero. But something thrown downward has a nonzero initial velocity.

AnnaSuppose an object is dropped from twenty meters, starting from rest, with downward positive. To find its falling time, use displacement equals initial velocity times time plus one half acceleration times time squared. Since the initial velocity is zero, twenty equals one half times nine point eight times time squared. That gives about two point zero two seconds.

MarcoThen use final velocity equals initial velocity plus acceleration times time. That gives a speed of about nineteen point eight meters per second just before impact, directed downward. Small differences in the last digit can come from rounding.

AnnaWhat if instead we know the impact speed? If an object starts from rest and reaches twenty-nine point four meters per second downward, that equation gives three seconds of falling.

MarcoAnd then the displacement equation gives forty-four point one meters downward. That’s the displacement from the release point, not necessarily the height above the ground. That distinction is easy to miss.

AnnaSo the practical habits are: identify whether air resistance is being ignored, choose a positive direction, use consistent signs, and select an equation with the quantities you know and the one you need.

MarcoAnd don’t treat nine point eight as a speed, mix sign conventions, assume every object starts from rest, or confuse distance fallen with height remaining. Those checks make the equations much easier to use.