Fluid mechanics

(i) Vertical speed of water jet on leaving nozzle = square-root[2g x 20] m/s
= 19.81 m/s (Take g, the acceleration due to gravity, equals to 9.81 m/s^2)

Time for a water particle to reach top of building = 20/(19.81/2) s = 2.019 s
Hence, horizontal speed of water jet = 10/2.019 m/s = 4.953 m/s
Therefore, velocity of water jet
= square-root [ 19.81^2 + 4.953^2] m/s = 20.42 m/s
Angle of nozzle with horizontal = arc-tan(19.81/4.953) degrees = 76 degrees

(ii) Let V be the water velocity in the hose. By equation of continuity,
(pi)(75/2)^2 x V = (pi).(25/2)^2 x 20.42
V = 2.269 m/s

Apply Bernoulli's equation at the hose and nozzle,
(P + Po) + (1/2)p(2.269^2) = Po + (1/2)p(20.42^2)
where Po is the atmospheric ressure
P is the water pressure in excess of atmospheric pressure
p is the density of water (= 1000 kg/m^3)

Hence, P = (1000/2)[20.42^2 - 2.269^2) Pa = 2.059 x 10^5 Pa
i.e. P = 206 kPa