A survey drone flies in a straight horizontal line 40 m above flat ground at a steady 15 m/s…

Step 1. The distance. $r = \sqrt{30^2 + 40^2} = 50\ \text{m}$.

Step 2. Its rate. From $r^2 = x^2 + 1600$ we get $2r\,\dfrac{dr}{dt} = 2x\,\dfrac{dx}{dt}$, so $\dfrac{dr}{dt} = \dfrac{x}{r}\,\dfrac{dx}{dt} = \dfrac{30}{50} \times 15 = 9\ \text{m/s}$.

Step 3. The angle. $\tan\theta = \dfrac{40}{x}$ gives $\sec^2\theta\,\dfrac{d\theta}{dt} = -\dfrac{40}{x^2}\,\dfrac{dx}{dt}$. Here $\sec^2\theta = \dfrac{r^2}{x^2}$. So $\dfrac{d\theta}{dt} = -\dfrac{40}{r^2}\,\dfrac{dx}{dt} = -\dfrac{40}{2500} \times 15 = -0.24\ \text{rad/s}$.

The friend's claim. The distance grows at $9\ \text{m/s}$ and not $15\ \text{m/s}$ at the moment given. When the drone is very far away, $\dfrac{x}{r}$ gets close to $1$, and the distance grows at nearly $15\ \text{m/s}$.

Why the distance grows more slowly. The drone does not move along the line of sight. Only the part of its velocity along that line, $\dfrac{30}{50} \times 15$, changes the distance, and the rest turns the line.