Water, ejected from the jet of a fire fighting engine, hits a wall perpendicularly at the rate of 12.2 m`cdot s^(-1)`. Assuming that the water does not recoil from the wall, calculate the pressuer developed on the wall. Mass of 1 `m^(3)` of water `10^(3)` kg.

Explanation

Let the area of cross section of the jet be A.
Hence, volume of water ejected per second = `A xx v` and mass of water hitting the wall per second = `A xx v xx d`, d = density of water = `10^(3) kg cdot m^(-3)`.
Change of momentum, per second, of the water jet hitting the wall
` "A"vd xx (v - 0) = Av^(2)d = A xx 12.2 xx 12.2 xx 10^(3) `
which is the force on the wall.
Hence, pressure on the wall = `("force")/("area") = (A xx 12.2 xx 12.2 xx 10^(3))/(A)`
`" " = 1.49 xx 10^(5) N cdot m^(2)` .