A non-dimensional model of a glider flying to minimise flight-time in the absence of lift and drag is COS Y = u' ' U == - sin y {' = u cosy, n' = usin y where is the flight-path angle, u is the non-dimensional velocity, $ is the non-dimensional range, and ʼn is the non-dimensional altitude and' (prime) denotes differentiation with respect to 7, the non-dimensional time. The glider travels from § = 0, n = ho to § = L,nh₁, where ho > h₁ > 0, and, at the start of the motion, the glider is pointing vertically downwards so that y = −π/2. that u = ccosy for some constant c 0 and hence show that dy/dt = 1/c. 1 differential equations for dε/dy and dŋ/dy. the general solutions of the differential equations in (b). that the flight-path is given by a cycloid. is the constant c determined? (Hint for (d): let x = §, y = ho — ŋ and 0 = π+2y.) - @ dr. du cos Y и .du ** "/di dr/de ·sin du. cost = ㅎ → 42 cas * u. cost = £ и Hence u= c cost. αξ dz = ucosɣ and 畿 dn dł == 4 = cast = dz u cost = 4² cas σ u sin r utan r Ӣ using = u² tan'} cost .u. The general solution. for ds = u² is E-du²+c for dr. u² tan.. » dn. = u². tan & dr. bagating both sides) n = u² sec² + Cz where. C. & Cz are negal I being the flight-path, its independent upon 8 flight-path hence flight path is agclaid.. cos(4) For X.= For r(t-sint) = *-! ' (l-cast) y=h0-716-π+28 4-(4-4) E=7(π+28)-sin (π +28)] hon = r. [1-cos (π +28)] = r. c.1+. cos 28) .n. = ho- r.CH+ cas 28). n = ho- E π1 + 28 + sin 28 (I+ cps 28) how to think to Create this 33. It Cos 28