Question

Eddie the Eagle, British Olympic ski jumper, is attempting his most mediocre jump yet. After leaving the end of the ski ramp, he lands downhill at a point that is displaced 54.4 m horizontally from the edge of the ramp. His velocity just before landing is 27.0 m/s and points in a direction 40.0$^\circ$ below the horizontal. Neglect any effects due to air resistance or lift.

a-What was the magnitude of Eddie's initial velocity as he left the ramp?
b-Determine Eddie's initial direction of motion as he left the ramp, measured relative to the horizontal.
c-Calculate the height of the ramp's edge relative to where Eddie landed.

Answer 1

Part a)

`v_f = 22.3 m/s`

Part b)

`\theta = 22.1 degree`

Part c)

`d = 11.7 m`

Step-by-step explanation:

Velocity just before it strike the ground is given as

`v_x = 27 cos40`

`v_x = 20.7 m/s`

`v_y = 27 sin40`

`v_y = 17.36 m/s`

since there is no friction in horizontal direction so its speed in horizontal direction will remain same

Part a)

velocity in X direction

`v_x = 20.7 m/s`

time taken by the skier to reach the ground is given as

`v_x t = \Delta x`

`20.7 t = 54.4`

`t = 2.63 s`

now in the same time it will cover vertical distance

`v_y = v_(oy) + at`

`-17.36 = v_(oy) + (-9.81)(2.63)`

`v_(oy) = 8.42 m/s`

so magnitude of initial speed is given as

`v_f = √(v_x^2 + v_y^2)`

`v_f = √(20.7^2 + 8.42^2)`

`v_f = 22.3 m/s`

Part b)

Direction of velocity

`tan\theta = (v_(oy))/(v_x)`

`tan\theta = (8.42)/(20.7)`

`\theta = 22.1 degree`

Part c)

Now in order to find the height of the ramp we can find the vertical displacement

`v_f^2 - v_y^2 = 2 a d`

`17.36^2 - 8.42^2 = 2(9.81)d`

`d = 11.7 m`