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 53.0 m horizontally from the edge of the ramp. His velocity just before landing is 26.0 m/s and points in a direction 37.0$^\circ$ below the horizontal. Neglect any effects due to air resistance or lift.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)

$v = 22.76 m/s$

Part b)

$\theta = 24.22 degree$

Part c)

$y = 8.04 m$

Explanation:

As we know that final velocity of the ski when it lands will be given as

$v_f = 26 m/s$ at an angle of 37 degree

so we will have

$v_{fx} = 26 cos37 = 20.76 m/s$

also in y direction the speed is given as

$v_{fy} = 26 sin37 = 15.65 m/s$

now we know that the displacement of the ski in horizontal direction is given as

$x = v_x t$

$53 = 20.76 t$

$t = 2.55 s$

so initial velocity in y direction is given as

$v_{fy} = v_y + at$

$-15.65 = v_y + (-9.8)(2.55)$

$v_y = 9.34 m/s$

So magnitude of initial velocity is given as

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

$v = \sqrt{9.34^2 + 20.76^2}$

$v = 22.76 m/s$

Part b)

for initial direction of motion we know that

$\theta = tan^{-1} \frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{9.34}{20.76}$

$\theta = 24.22 degree$

Part c)

For height of the ramp we know that

$v_{fy}^2 - v_y^2 = 2 a d$

$15.65^2 - 9.34^2 = 2(9.81)(d)$

$y = 8.04 m$