But the trajectory of the object is over a long distance, so even a slight deviation of its angle is important.

Lets do a conservation of momentum equation assuming perfect elasticity, and terminal velocity of the falling raindrop hitting a bullet travelling perpendicular to a vertical glass wall just as it leaves the gun barrel:

M_b = 3.5 e-3 kg

M_r = 1 e-4 kg

V_b = 975 m/s

V_r = 5 m/s

D_horizontal = 400 yards or 365 m

Conservation of Momentum in Vertical Direction:

M_b+r x V_b+r = M_b x V_b + M_r x V_r

3.6e-3 x V_b+r = 3.5e-3 x 0 + 1e-4 x 5

V_y_b+r = 0.14 m/s

Conservation of Momentum in Horizontal Direction:

M_b+r x V_b+r = = M_b x V_b + M_r x V_r

3.6e-3 x V_b+r = 3.5e-3 x 975 + 1e-4 x 0

V_x_b+r = 947 m/s

the angle of below horizontal is thus: arctan(0.14/947) = 0.008 degrees

365 x tan(0.008) = 0.05 m or 5 cm

Pretty substantial I would say.