(1)

P = G ω cos δ(v cos α − u cos δ)2;
g cos α

(2)

Pu = Gωu cos δ(v cos α − u cos δ)2;
g cos α

(3)

Three cases may be distinguished:—

(a) The plane is at rest. Then u = 0, N = (G/g) ωv2cos α; and the work done on the plane and the efficiency of the jet are zero.

(b) The plane moves parallel to the jet. Then δ = α, and Pu = (G/g)ωu cos2 α (v − u)2, which is a maximum when u = 1⁄3v.

When u = 1⁄3v then Pu max. = 4⁄27(G/g)ωv3 cos2α, and the efficiency = η = 4⁄9cos2α.

(c) The plane moves perpendicularly to the jet. Then δ = 90° − α; cos δ = sin α; and Pu = G/g ωu (sin α / cos α) (v cos α − u sin α)2. This is a maximum when u = 1⁄3v cos α.