A student measures the time taken for a stone to fall from a cliff....

First, find the mean value of the time measurements:
$t_{mean} = \dfrac{9.8 + 9.9 + 10.1 + 10.2}{4} = \dfrac{40.0}{4} = 10.0\text{ s}$.

Next, calculate the absolute errors for each reading:
$\Delta t_1 = |9.8 - 10.0| = 0.2\text{ s}$
$\Delta t_2 = |9.9 - 10.0| = 0.1\text{ s}$
$\Delta t_3 = |10.1 - 10.0| = 0.1\text{ s}$
$\Delta t_4 = |10.2 - 10.0| = 0.2\text{ s}$

The mean absolute error in time is:
$\Delta t_{mean} = \dfrac{0.2 + 0.1 + 0.1 + 0.2}{4} = \dfrac{0.6}{4} = 0.15\text{ s}$.

The percentage error in the time measurement is:
$\dfrac{\Delta t_{mean}}{t_{mean}} \times 100 = \dfrac{0.15}{10.0} \times 100 = 1.5\%$.

The height is calculated using $h = \dfrac{1}{2}gt^2$. Since $g$ and $\dfrac{1}{2}$ are exact constants, the relative error in $h$ depends only on $t$. According to the rule for propagation of errors for a quantity raised to a power:
$\dfrac{\Delta h}{h} \times 100 = 2 \times \left(\dfrac{\Delta t}{t} \times 100\right)$

Percentage error in $h = 2 \times 1.5\% = 3.0\%$.

Answer: $3.0\%$