Let the jet of water, coming out from the nozzle, strikes an inclined flat plate as shown in the figure

Let, v = Velocity of the jet in the direction of x,

$\theta$ = Angle between the jet and the plate,

a = Area of cross-section of the jet.

The mass of water per second striking the plate,

$=\rho\times av$

If the plate is smooth and if it is assumed that there is no loss of energy due to the impact of the jet it, then the jet will move over the plate striking with a velocity equal to the initial velocity.

Let us find the force exerted by the jet on the plate in the direction normal to the plate.

$F_n$ = mass of jet striking per second $\times$ [Initial velocity of the jet before striking in the direction of n - final velocity of the jet after striking in a direction of n]

$F_n=\rho av[v\sin\theta-0]$

$F_n=\rho av^2\sin\theta.............(1)$

This force can be resolved into two components, one in the direction of the jet and the other perpendicular to the direction of flow. Then we have,

$F_x=F_n\cos(90^\circ-\theta)$

$F_x=F_n\sin\theta$

$F_x=\rho A v^2\sin\theta\times\sin\theta \space \space [\because F_n=\rho av^2\sin\theta ] ........(2)$

And,

$F_y=F_n\sin(90^\circ-\theta)$

$F_y=F_n\cos\theta$

$F_y=\rho Av^2\sin\theta\cos\theta................(3)$