Question

You’re a consulting engineer specializing in athletic facilities, and you’ve been asked to help design the Olympic ski jump pictured in figure below. Skiers will leave the jump at 26 m/s and 9.5∘ below the horizontal, and land 55 m horizontally from the end of the jump. Your job is to specify the slope of the ground so skiers’ trajectories make an angle of only 3.0∘ with the ground on landing, ensuring their safety.(Figure 1)

Answer 1

Given the initial velocity, U = 26 m/s

Thus Ux = 26 cos9.5° = 25.65 m/s

and Uy = 26 sin9.5° = 4.29 m/s

Using, R = [Ux] x t

=> 55 = 25.65 x t

=> t = 2.14 s

Thus by v = u + gt

=> Vy = Uy + gt

=> Vy = 4.29 + 9.8 x 2.14

=> Vy = 25.26 m/s

Thus by tanθ = Vy/Vx {Vx = Ux = constant}

=> tanθ = 25.26 / 25.65 = 0.9848

=> θ = tan^-1(0.9848) = 44.56°

Thus to have a 3* landing the trajectory of plane will be (44.56° - 3°) = 41.56° elevation from horizon.