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If you look at the standard Net Take-off flight path, the speed an "airliner" should achieve at the screen height is the V2 speed, a speed that ensures a minimum climb gradient in case of an engine failure. For a twin jet this should be a minimum of 2.4% climb gradient in case of engine failure until acceleration altitude. However if you look at the standard SID gradient, which is composed of the following: starting from 5m above DER - Departure End of Runway it is designed to have a minimum of 2.5% gradient + 0.8% clearance on top of that. So the minimum standard SID gradient = 3.3%. So this means that the airplane w/ V2 and an engine failure will under perform and likely hit obstacles no? (In visual conditions you can visually steer away, but in IMC conditions this is not possible?)So here I am not sure what is wrong in my logic? or is there something I am confusing with? Thanks in advance!

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