Fluid mechanics

(i) Because pressure at valve exit = (0.5)pg = (1/2)p.v^2
where p is the density of coffee
g is the acceleration due to gravity (= 9.81 m/s^2)
v is the speed of coffee at the exit
Hence, v = square-root [ 2g x 0.5] m/s = 3.132 m/s

(ii) Assume the coffee is under free fall, use equation of motion v^2 = u^2 + 2a.s
where u = 3.132 m/s, a = g(=9.81 m/s^2), s = 0.5 m
hence, v^2 = (3.132^2 + 2 x 9.81 x 0.5) (m/s)^2
v = 4.429 m/s

Use equation of continuity at the valve exit and the jet near ground,
(pi).(1/2)^2 x 3.132 = (pi)(d/2)^2 x 4.429
where d is the diameter of coffee jet near ground
hence, d = square-root [3.132/4.429] cm = 0.8409 cm