Let's find time of flight from the descent:
With y axis pointing upwards, Δy = -0.75m , starting velocity vy0= 0, acceleration a=-g= -9.8 m/s2
Δy = vy0t + (1/2) at2 = (1/2) at2 i.e. t =sqrt(2Δy/a) =sqrt( 2*0.75/9.8) =0.39 s
During this time it undergoes a horizontal displacement Δx=20m at a constant horizontal velocity vx0
Δx = vx0 t => Exit velocity, vx0 = Δx /t = 20m /0.39s = 51 m/s [ two significant digits]
Assuming constant acceleration within the barrel, a= (v2-v02)/2Δx gives
a = 512/(2*1.15) = 1130m/s2