Answer

(1/2)ρ(V22-V12) + ρg(z1-z2) + P2 - P1 = 0

(1/2)ρ(V22-V12) + P2 - P1 = 0

P1 = (1/2)ρ(V22-V12) + P2

P1 = 0.5*998*(11,3002-(5.46*10-6)2) + 101,325

P1 = 6.34*1010Pa

Woooo you did it! Well done :)))

Consequently, it is theoretically possible for a water gun to shoot the moon, but only if the pressure is 6.34*1010Pa, or 626 atmospheres. Surprisingly, this is highly unrealistic, as the average pressure in a water gun is 3 atmospheres.