Calculus
# Arc Length and Surface Area

A delivery drone flying at constant speed \(15 \text{ m/s}\) and constant height \(2700 \text{ m}\) toward a destination drops its goods. If the trajectory of the falling goods until it hits the ground can be described by the equation
\[y=2700-\frac{x^2}{75},\]
where \(x\) is the horizontal distance it travels and \(y\) is its height above the ground, what is the **distance** (not horizontal displacement) traveled by the goods until it hits the ground?

**Note:** You can use \(\displaystyle \int \sqrt{a^2+u^2}\,du=\frac{u}{2}\sqrt{a^2+u^2}+\frac{a^2}{2}\ln(u+\sqrt{a^2+u^2})+C.\)

(i) over any specified interval, the area between \(f(x)\) and the \(x\)-axis is equal to the arclength of the curve, and

(ii) \(f(0) = 1\).

If \(S = \displaystyle\int_{-1}^{2} f(x) dx\), then find \(\lfloor 1000S \rfloor\).

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