All Four Kinematic Equations

When solving free fall problems, the third kinematic equation $v^2 = v_0^2 + 2a\Delta y$ becomes invaluable when time is unknown or unnecessary. Since acceleration in free fall is $a = -g$ (taking upward as positive), the equation becomes:

$$v^2 = v_0^2 - 2g\Delta y$$

where $\Delta y = y - y_0$ is the vertical displacement.

Common Free Fall Patterns

Objects dropped from rest: When an object is dropped from height $h$ (so $y_0 = h$, $v_0 = 0$) and falls to the ground ($y = 0$), we have $\Delta y = -h$. The impact speed is:

$$v^2 = 0 - 2g(-h) = 2gh$$

Therefore $v = \pm\sqrt{2gh}$. Since the object is moving downward at impact, we choose the negative sign if upward is positive: $v = -\sqrt{2gh}$.

Minimum launch speed to reach a height: To find the minimum upward launch speed $v_0$ needed to reach height $h$ above the launch point, we use the fact that at maximum height, $v = 0$:

$$0 = v_0^2 - 2gh$$

Solving: $v_0 = \sqrt{2gh}$

General case: For any free fall problem where an object starts at $y_0$ with initial velocity $v_0$ and moves to position $y$, the final velocity is:

$$v = \pm\sqrt{v_0^2 - 2g(y - y_0)}$$

The sign depends on the direction of motion at that point.

Quick Numerical Examples

Using $g = 9.8$ m/s²:

Drop from 80 m: An object dropped from rest falls 80 m. Its impact speed is $v = \sqrt{2 \cdot 9.8 \cdot 80} = \sqrt{1568} \approx 39.6$ m/s downward.

Reach 50 m height: To throw an object up to 50 m requires initial speed $v_0 = \sqrt{2 \cdot 9.8 \cdot 50} = \sqrt{980} \approx 31.3$ m/s upward.

Combined motion: An object at $y_0 = 20$ m is thrown upward with $v_0 = +15$ m/s. When it hits the ground ($y = 0$):

$$v^2 = 15^2 - 2(9.8)(0 - 20) = 225 + 392 = 617$$ $$v = -\sqrt{617} \approx -24.8 \text{ m/s}$$

The negative sign indicates downward motion at impact.