Given P=$P_{1}=P_{2}=39.24N/cm^{2}=39.24\times 10^{4} N/m^{2}$

Q=250 d/s = $0.25m^{3}/s$

d=$d_{1}=d_{2}$=300mm=0.3 m

A=$A_{1}=A_{2}=\frac{\pi}{4}\times d^{2}=\frac{\pi}{4}\times (0.3)^{2}=0.67069m^{3}$

Q=AV

Velocity of water V=$V_{1}=v_{2}=\frac{Q}{A}=\frac{0.05}{0.077069}$=3.537m/s

From impulse-momentum principle, the force on x direction

F=$P_{1}A_{1}-(-P_{2}A_{2}cos \theta)+PQ[V_{1}-(-V_{2}Cos \theta)$

=PA$(1+cos\theta)+PQV(1+cos\theta)$

=$(39.24\times 10^{4})\times 0.07069(1+cos45)+1000\times 0.25\times 3.537(1+cos45)$

=48863 N

Similarity the force along y-direction using momentum equation can be calculated as

o-$P_{2}A_{2}sin\theta-Fy=PQ[V_{2}sin\theta-0]$

Fy=$-P_{2}A_{2}sin\theta-PQVsin\theta=$

=$-(P_{2}A_{2}+PQV_{2})sin\theta$

=-$[(39.24\times 10^{4})\times 0.07069+1000\times 0.25\times 3.537]$sin 45

=-20240N

1. Magnitude of resultant Force, F

F=$\sqrt{F_{x}^{2}+F^{2}_{y}}=\sqrt{(48.869)^{2}+(-26240)^{2}}$

=52888.8N

1. Direction of resultant for $\Phi$

$Q=tan^{-1}(\frac{Fy}{Fx})=\tan^{-1}(\frac{-26240}{48863})=-22.5^{\circ}$