I think you want to determine the exit speed?
You have to determine how much velocity was decreased by calculating it from the kinetic energy.
[tex]KE = \frac{1}{2} mv^{2} [/tex]
[tex]1.4 x 10^{5}= \frac{1}{2} (1100) v^{2} [/tex]
[tex] v^{2} =254.55[/tex]
[tex]v = 15.95 m/s[/tex]
So the velocity reduces by 15.95 m/s. Subtracting this to the initial velocity: 22 - 15.95 = 6.05 m/s.
So, the final speed was 6.05 m/s.
I hope I was able to help :)
