Free Fall and the Equations of Motion

The question

You drop a stone into a well and hear the splash 2.0 s later. How deep is the well?

About 19.6 m. Starting from rest and gaining 9.8 m/s every second, the stone is moving at 19.6 m/s after 2.0 s, and the distance it falls is the area under its velocity–time graph: ½ × 2.0 s × 19.6 m/s = 19.6 m — the same as $s=\frac{1}{2}a{t}^2$. The mass of the stone makes no difference. If we also allow for the time the sound takes to come back up, the well is about 18.5 m deep.

In words, then in symbols

• The velocity at the end is the velocity at the start plus the acceleration multiplied by the time.

  $v=u+at$

• The displacement is the starting velocity times the time, plus half the acceleration times the time squared.

  $s=ut+\frac{1}{2}a{t}^2$

• The final velocity squared is the starting velocity squared plus twice the acceleration times the displacement.

  $v^2=u^2+2as$

How to use the equations of motion

1. Choose a positive direction.
2. List what you know — $u$, $v$, $a$, $s$, $t$ — each with its sign.
3. Pick the equation that contains your unknown and no other unknown.
4. Substitute, signs included, and solve.
5. Check the answer against the velocity–time graph.