When you drop a ball, its position changes in a complicated way. How would you calculate that position?
When you drop a ball, its altitude decreases by larger and larger increments as the seconds pass. If we call the altitude from which you drop it zero, then its altitude after 1 second is -4.9 m (-4.9 meters or about -16 feet), after 2 seconds is -19.6 m, and after 3 seconds is – 44.1 m. Here is one way to calculate those values.
First, note that the ball is accelerating downward steadily at 9.8 m/s2. The ball’s initial velocity was zero, so its velocity after falling for time t is 9.8 m/s2* t downward.
Next, let’s find the ball’s average velocity while falling for time t. The ball’s velocity has been changing steadily from 0 when you dropped it to 9.8 m/s2* t downward after falling for time t, so it’s average velocity is simply the average of those two individual values: 0 and 9.8 m/s2* t downward. That average is 4.9 m/s2 downward.
Lastly, let’s determine how far downward the ball has traveled after falling for time t. Since it’s average velocity was 4.9 m/s2* t downward and it has traveled for time t with that average velocity, its change in position is 4.9 m/s2* t downward * t or simply 4.9 m/s2* t2 downward. As you can see, its change in position is proportional to the square of its fall time t. With each passing second, it is moving downward faster and covering more distance. As stated above, its altitude after 1 second of falling is -4.9 m, after 2 seconds of falling is -19.6 m, and after 3 seconds of falling is -44.1 m.