Difference between theoretical and experimental time due to gravity This is a very rudimentary question, but I thought I would have to ask here because I don't discern any site better.

Given the formula that $t$ equals the square root of $2h/g$, calculate the theoretical and experimental time needed for an object to reach the ground. For the experimental time, add a random amount of error of $0$ to $0.05$ seconds to the theoretical time.

I don't comprehend the concept behind this. What is the difference between the experimental and theoretical time, and how does one add the random amount of error?
Thanks in advance.
 A: Suppose you have an object at rest ($v_{iy}$) and at a height $y_{i}$ (otherwise designated as $h$) and you allow the object to be dropped at a time $t_{i} = 0$ until it hits the ground. We will allow for the ground in our coordinate system to be the origin along the y-axis. As a result, $y_{f} = 0$, and therefore $\Delta{y} = -y_{i} = h$
From there, we can derive that indeed:
$$t=\sqrt{\frac{2h}{g}}$$
Until this point, our theoretical calculations are spot-on. This is however, solely within the mathematical framework of Newtonian mechanics. In order to make the result more applicable to real life, we simulate imprecisions in systems, which in this case could include things such as the the reflex time between starting the timer and dropping the ball, the act of approximating gravity ($g$), the imprecision in measuring the height ($h$), and other unforseen variables.
From what it looks like, it's asking you to take your result $t$, whatever it may be, and turn it into something of the form $t\pm{p}$, where $\pm{p}$ represents your acknowledgement of those errors and imprecisions, say $t\pm{0.03s}$.
