An intense stream of water of cross-section area $A$ strikes a wall at an angle $\theta$ with the normal to the wall and returns back elastically. if the density of water is $\rho$ and its velocity is $v$, then the force exerted on the wall will be:

Here, we need to find the energy used by the water stream in the wall. The term elasticity and velocity have already indicated that we will apply the change to the pressure formula.
Let's imagine the question in the form of a drawing.

We know that the energy used in the body is nothing but a change in the intensity of that body.
$F =P\Delta t ---- \left(i\right)$
Here, power in a stream is nothing but a change in its intensity. And the stream of water uses the same force on the wall. Thus, the energy used in a wall is useless without a change in water pressure.
Here, we need to get power per second. Therefore, the above formula (i) can be changed to:
$F = \Delta P ---- \left(ii\right)$
We know, the size of a stream of water per second is given by:
Weight = volume × density
$m = V \times \rho$
Here, the flow rate of water per second = Location × speed = A × v
Thus, mass = Location × speed × density
$\Rightarrow m = A \times v \times \rho ---- \left(iii\right)$
Now,
$F = \Delta P = Pf - Pi$
Thus, $F=m{v}_f-m{v}_i$
Here, the energy is applied to the wall only because of the part perpendicular to the wall because it is this part that will use the energy in the wall. This is part of the cosmos of momentum. Part of the sin is along the wall so it does not use force on the wall. It can therefore be ignored.
Thus, the strength of the wall is given as follows:
$F = m{v}_f\cos \theta - (- m{v}_i\cos \theta )$
Also, the conflict here is natural. So the first and last speeds of the stream will always be the same.
So,
$v_f = v_i = v \left(stop\right)$
Then, $F = 2mv\cos \theta$
Using, equation (iii), we replace m:
Thus, $F = 2A{v}^2\rho \cos \theta$
Therefore, the appropriate option is (c).