<s>[INST] Average height of raindrops from height $h$ 
Raindrops drip from a spout at the edge of a roof and fall to the ground. Assume that the drops drip at a steady rate of $n$ drops per second (where $n$ is large) and that the height of the roof is $h$.
(c) What is the average height of these raindrops as $n$ grows larger when the first drop hits the ground?

First we need to find the time that the first raindrop hits the ground. We know from the projectile motion formula that
$$-\frac{1}{2}gt^2+h=0\Longrightarrow t=\sqrt{\frac{2h}{g}}$$
We then substitute it back into the first equation:
$$-\frac{1}{2}g{\left(\sqrt{\frac{2h}{g}}\right)}^2+h=0.$$
Then we solve for $h$:
$$h_{\text{avg}}=\frac{1}{2}g{\left(\sqrt{\frac{2h}{g}}\right)}^2$$
We add an additional factor since to find the limit of the average as n gets large. We get
$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\frac{1}{2}g{\left(\frac{i}{n}\sqrt{\frac{2h}{g}}\right)}^2=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\frac{i^2}{n^2}{h}.$$
Ultimately we solve
$$h\lim_{n\to\infty}\sum_{i=1}^{n}\frac{i^2}{n^2}\frac{1}{n},$$
which can be converted into
$$h\int_{0}^{1}x^2\,\mathrm dx=\frac{1}{3}h.$$
(This is the distance from the roof down. The real height is $\dfrac{2}{3}h$.)
Am I correct? If not, how do I find the average height?
 [/INST] $\def\peq{\mathrel{\phantom{=}}{}}$Assume instead that there are $n$ drops per $T$ seconds so that $n$ becomes dimensionless, thus the time interval between two drops is $Δt = \dfrac{T}{n}$.
It takes $t_1 = \sqrt{\dfrac{2h}{g}}$ for the first drop to hit the ground, and when it hits, there are $N = \left[ \dfrac{t_1}{Δt} \right] + 1$ drops in the air (including the first drop) and the $k$-th drop has traveled $t_k = t_1 - (k - 1)Δt$, which implies its current height is $h - \dfrac{1}{2} g t_k^2$. Therefore, the average height is\begin{align*}
&\peq \frac{1}{N} \sum_{k = 1}^N \left(h - \frac{1}{2} g t_k^2 \right) = h - \frac{g}{2N} \sum_{k = 1}^N (t_1 - (k - 1)Δt)^2\\
&= h - \frac{g}{2N} \left( N t_1^2 - 2t_1 Δt \sum_{k = 1}^N (k - 1) + (Δt)^2 \sum_{k = 1}^N (k - 1)^2 \right)\\
&= h - \frac{g}{2N} \left( N t_1^2 - N(N - 1) t_1 Δt + \frac{1}{6} (2N - 1)N(N - 1) (Δt)^2 \right)\\
&= h - \frac{1}{2} g t_1^2 + \frac{1}{2} (N - 1) g t_1 Δt - \frac{1}{12} (2N - 1)(N - 1) g(Δt)^2\\
&= \frac{1}{2} (N - 1) g t_1 Δt - \frac{1}{12} (2N - 1)(N - 1) g(Δt)^2.
\end{align*}
Note that $N Δt → t_1$ as $n → ∞$, thus\begin{align*}
&\peq \lim_{n → ∞} \left( \frac{1}{2} (N - 1) g t_1 Δt - \frac{1}{12} (2N - 1)(N - 1) g(Δt)^2 \right)\\
&= \frac{1}{2} g t_1^2 - \frac{1}{6} g t_1^2 = \frac{2}{3} h.
\end{align*}
</s>