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Old 14th June 2011 | 00:04
  #1970 (permalink)  
Litebulbs
 
Joined: Oct 2001
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From: Gatwick
JD-EE

Originally Posted by JD-EE
Kinetic energy is 1/2 mV^2. Potential energy is mgh. So the mass washes out of the equation. So a 3000' climb would give "gh" = 1/2 v^2 = 96000 ft^2/s^2 equals about a 438 '/s velocity change or about 300 MPH.
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?
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