Answer :
To develop this problem we will apply the energy conservation theorem and the principle of work. Basically we will have that the kinetic, potential and work energy must be equivalent to the kinetic energy just before the impact. We will start converting the units given to the British system and then proceed with the Calculations.
Remember that according to the energy balance in this case it would be balanced like this
[tex]T_1 +\sum U_{1-2} = T_2[/tex]
[tex]\frac{1}{2} mv_1^2+mgh -W_{drag} = \frac{1}{2} mv_2^2[/tex]
Here
m = mass
[tex]v_{1,2}[/tex]= Velocity at each moment
[tex]W_{drag}[/tex]= Work by drag
h = Height
g = Acceleration due to gravity
Mass
[tex]m =255000 lb (\frac{1slug}{32.174lb})[/tex]
[tex]m = 7925.654slug[/tex]
Initial Velocity
[tex]v_1 = 550 \frac{mi}{h} (\frac{5280ft}{1mi})(\frac{1hr}{3600s})[/tex]
[tex]v_1 = 806.667ft/s[/tex]
Work by drag
[tex]W_{drag} = (2.96*10^6BTU)(\frac{778.169lb\cdot ft}{1BTU})[/tex]
[tex]W_{drag} = 2303380240lb\cdot ft[/tex]
By work energy principle
[tex]\frac{1}{2} mv_1^2+mgh -W_{drag} = \frac{1}{2} mv_2^2[/tex]
Replacing,
[tex]\frac{1}{2} (7925.654slug)(806.667ft/s)^2 +(7925.654slug)(32.08ft/s^2)(30000ft) -2303380240lb\cdot ft = \frac{1}{2} (7925.654slug)v_2^2[/tex]
Solving for [tex]v_2[/tex], we have that
[tex]v_2 = 1412.2 ft/s[/tex]
Converting this value,
[tex]v_2 = 1412.2 ft/s (\frac{1mi}{5280ft})(\frac{3600s}{1h})[/tex]
[tex]v_2 = 962.85mi/h[/tex]
Therefore the velocity of the aircraft at the time of impact is 962.85mi/h