Resort site on hill

After review, the engineer decided it would be better to place the new resort (point A) further up the hill (see origianl case). The horizontal distance from the front of the hill is 200 ft. Determine the pressure at the proposed site for a wind speed of 50 mph.

Assume the hill can be approxiatee by a half-body rankine shape. The hill has a maximum height of 300 at infinite. Use air density as 0.002377 slug/ft³. Assume the up stream pressure is 0 gage pressure.

Need to find the location of of the resort (point A) in terms of θ and r. Then, the velocity at A can be determined. Half-body rankine shape is given by
    b (π - θ) / sinθ = r
    b (π - θ) = r sinθ = y

The body height goes to 300 ft as θ goes to zero
    b (π - 0) = 300
    b = 300 / π = 95.49 ft

Thus, in terms of r and θ, the body shape is
     95.49 (π - θ) = r sinθ

The resort, A, is 200 ft from the stagnation point,
     b + r cosθ = 200
     95.49 + r cosθ = 200
    r = 104.5 / cosθ

Put this into the body shape equation gives,
     95.49 (π - θ) = (104.5 / cosθ) sinθ
   0.9138 (π - θ) = tanθ

This is a non-linear equation that cannot be solve with algebra. Using WolframAlpha.com ("0.9138 (3.1416 - x) - tan(x) = 0", the smallest root for θ is

    θ = 1.0823 rad = 60.01 deg

Thus r to point A is
     95.49 (π - 1.0823) = r sin60.01
    r = 227.0 ft

The upstream velocity is 50 mph, which is 73.33 ft/s. The final velocity at A is
    
        = 5,377 (1 + 0.1770 + 0.4999)
    V_A² = 9,017 ft²/s²

Pressure at A is found using Bernoulli's equation
   p_o + 0.5ρU² + ρgz_o = p_A + 0.5ρV_A² + ρgz_A
   0 + 0.5 (0.002377) 73.33² + 0 = p_A +
              0.5 (0.002377) 9,017 +
                        0.002377 (32.2) (227.0 sin60.01)
   6.391 = p_A + 10.727 + 15.048

p_A = -19.38 lb/ft² = -0.1045 psi
         (below ATM gage pressure)