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

Answer:

Part a)

$v_f = 22.3 m/s$

Part b)

$\theta = 22.1 degree$

Part c)

$d = 11.7 m$

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

$v_f = \sqrt{v_x^2 + v_y^2}$

$v_f = \sqrt{20.7^2 + 8.42^2}$

$v_f = 22.3 m/s$

Part b)

Direction of velocity

$tan\theta = \frac{v_{oy}}{v_x}$

$tan\theta = \frac{8.42}{20.7}$

$\theta = 22.1 degree$

Part c)

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

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

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

$d = 11.7 m$