If it strikes the ground with an impact velocity of $40\ \mathrm{\displaystyle \frac{m}{s}}$, what acceleration did it experience during its descent?

$a=10\ \mathrm{\displaystyle \frac{m}{\ s^2}}$

If something is dropped, then that means it has no initial velocity: $v_0 = 0$. So, we’re given $v_0$, $\Delta x$, and $v$, and we’re asked for $a$. Since $t$ is missing, we use Big Five #5:

$$v^2=v^2_0+2a(x-x_0)$$

$$v^2=2a(x-x_0)$$

(since $v_0=0$)

$$a=\frac{v^2}{2(x-x_0)}$$

$$a=\frac{\left(40\ \mathrm{\displaystyle \frac{m}{s}}\right)^2}{2(80\ \mathrm{m})}$$

$$a=10\ \mathrm{\displaystyle \frac{m}{\ s^2}}$$

Note that since a has the same sign as $(x – x_0)$, the acceleration vector points in the same direction as the displacement vector. This makes sense here, since the object moves downward, and the acceleration it experiences is due to gravity, which also points downward.