# Related rates problem, rocket and observer

I have been leafing through the other related rate problems posted here, so I hope this is not a duplicate.

The problem (from the 3,000 series) is as follows: A rocket is shot very vertically upward with an initial velocity of 400 ft/s. Its height s after t seconds is $s=400t-16t^{2}$ How fast is the distance changing from the rocket to an observer on the ground 1800 ft away from the launching site, when the rocket is still rising and is 2400 feet above the ground?

Part of the answer states the following, which I do not understand:

When $s=2400, u^{2}=(100)^2*(900)$ ($u$ being the distance from the rocket to the observer).

Can anyone explain this to me? Where are these numbers from?

$u=\sqrt {1800^2+2400^2}$
$u^2=100^2*(18^2+24^2)$
$u^2=100^2*(900)$