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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.

Respuesta :

Answer:

Part a)

[tex]v_f = 22.3 m/s[/tex]

Part b)

[tex]\theta = 22.1 degree[/tex]

Part c)

[tex]d = 11.7 m[/tex]

Explanation:

Velocity just before it strike the ground is given as

[tex]v_x = 27 cos40[/tex]

[tex]v_x = 20.7 m/s[/tex]

[tex]v_y = 27 sin40[/tex]

[tex]v_y = 17.36 m/s[/tex]

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

Part a)

velocity in X direction

[tex]v_x = 20.7 m/s[/tex]

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

[tex]v_x t = \Delta x[/tex]

[tex]20.7 t = 54.4[/tex]

[tex]t = 2.63 s[/tex]

now in the same time it will cover vertical distance

[tex]v_y = v_{oy} + at[/tex]

[tex]-17.36 = v_{oy} + (-9.81)(2.63)[/tex]

[tex]v_{oy} = 8.42 m/s[/tex]

so magnitude of initial speed is given as

[tex]v_f = \sqrt{v_x^2 + v_y^2}[/tex]

[tex]v_f = \sqrt{20.7^2 + 8.42^2}[/tex]

[tex]v_f = 22.3 m/s[/tex]

Part b)

Direction of velocity

[tex]tan\theta = \frac{v_{oy}}{v_x}[/tex]

[tex]tan\theta = \frac{8.42}{20.7}[/tex]

[tex]\theta = 22.1 degree[/tex]

Part c)

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

[tex]v_f^2 - v_y^2 = 2 a d[/tex]

[tex]17.36^2 - 8.42^2 = 2(9.81)d[/tex]

[tex]d = 11.7 m[/tex]

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