I am probably being a bit thick, but does this relate to the aircraft in question and if so, are you saying the aircraft lost 300mph in its climb?
JD-EE:
Roughly speaking - I cheated on the math somewhat. Rudderrat posted the more correct equation from which mass drops out. I presumed a drop to zero velocity which is a cheat. It gives the general idea, though. A plane drops a staggering amount in the 3000' climb with no additional power input through the energy tradeoff.
For those having a hard time beleiving a massive (true) speed drop from the climb consider a cart on a roller coaster going fast at the bottom of the first drop, as it reaches the top of the next hill it's velocity drops to close to 0 then accelerates again as it goes down.
The a roller coaster cart has no thrust and less friction/drag losses than an airplane however the principle is the same: trade speed for altitude.
Note, ignoring drag no energy is "lost" just converted from kinetic to potential, at the top of climb most of it is still available to convert to speed if the nose is pointed down.