Why still no dive acceleration difference?

[BlackBerry]

We are going to just simplify the drag and thrust values to illustrate the mechanics of solving the problem.

Characteristics of our Airplane:

Weight 9000lbs
Thrust in lbs = 1000lbs
Drag in lbs = 500

Zoom climb from 300mph to Vy at a 45 degree angle:
The airplane will move to equilibrium

Entry speed = 300mph = 441fps
Zoom Angle 45 degrees
Vy = 150mph = 220.5fps

Zoom height:

Sum the forces on the flight path -

9000lbs * sin 45 = 6364lbs
1000lbs – 500lbs - 6364lbs = 5864lbs

a = F/m

m = 9000lbs/32.2 = 279.5 lb-s^2/ft
a= 5864lb/279.5lb-s^2/ft
a = 20.98 ft/s^2

s = (V1^2 – V2^2 ) / 2a

s = (441^2 – 220.5^2)/(2 * 20.98ft/s^2) = 3476.18 ft

3476.18 ft * sin 45 = 2458 ft

Crumpp，when airplane zoom 45 degree from 300mph to 150mph, It takes a quite long time: 20 seconds?

In the whole zoom process, Is the acceleration a constant or a variable ? All your formular is of "constant accelerated motion". That's not correct.

In fact, when airplane begin to zoom, the "a"=20.98ft/s^2, however, after a few seconds, as speed drops, the engine thrust increase to 1100lbs, and the air drag will be 450lbs, so the "a" is no long 20.98, "a" will be smaller than 20.98. The acceleration changes during the whole zoom process, so it's a "variable accelerated motion".

I am surprised, your math model is too simple to get correct result.