### 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/s^{2}. The ball’s initial velocity was zero, so its velocity after falling for time *t* is 9.8 m/s^{2} ** 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/s^{2} ** 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/s^{2} ** t* downward. That average is 4.9 m/s^{2} 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/s^{2} ** t* downward and it has traveled for time *t* with that average velocity, its change in position is 4.9 m/s^{2} ** t* downward * *t* or simply 4.9 m/s^{2} ** t ^{2}* 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.