[bonesmccoy]

OK...here's my verticle ballistics formula

Calculating time to impact based upon verticle acceleration of 32 feet per sec per sec


Previous data point is
13:56:24 GMT
36.48154, -110.4165
H= 220235 ft
V= 21.17 mach

In 6 seconds the vehicle went from 220235 to 219820 feet.
That's about 415 feet.

I'm going to assign an initial verticle acceleration of 0.
Final verticle acceleration of the object is equal to g = 32 feet/sec^2.

If
y = 219820 feet
v(iy) = 0 ft/sec
a(y) = 32 ft/sec/sec

-219820 ft = (0 ft/sec)(t) + 0.5*(-32 ft/sec/sec)*(t^2)

first term on right cancels due to zero, simplifying

-219820 ft = (-16 ft/sec/sec)*t^2

(-219820/-16) = 13738.75 = t^2

square root of 13738.75 = 117.2124 seconds

[John Jamieson]

I'm afraid aero drag plays a big role here! That's much to simple.

[bonesmccoy]

Horizontal Ballistics Calculation for object departing at Time of First Roll Reversal

x = v(ix)*t + 0.5*a(x)*t^2

Defined
v(ix) = initial velocity of object when departing vehicle
If v = Mach 21.13 at 13:46:30GMT,(46)1st Roll Reversal initiation
Assuming velocity= 21.13 * (761 mph) = 16079.93 mph
16079.93 mph * (1/3600 hour/sec) * (5280 ft/mile) = 23583.9 ft/sec

a(x) = 0 (is that correct guys?)

Time = 117.2124 seconds

x = (23583.9 ft/sec)*(117.2124 sec) + 0.5*(0 ft/sec/sec)*(117.2124 sec)^2

Since the left most factor drops out,
x = 2764326 feet = 523.5465 miles = 523 miles and 2885 feet

Of course, I am not accounting for any atmospheric factors, drag, etc.

I'll try and plot a map.