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)
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.
You’re a consulting engineer specializing in athletic facilities, and you’ve been asked to help design the...
Part A 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 27 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) What slope...