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zloy xaker [14]
3 years ago
5

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 72.1 m horizontally from the edge of the ramp. his velocity just before landing is 33.0 m/s and points in a direction 30.0$^\circ$ 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?
Physics
1 answer:
DedPeter [7]3 years ago
5 0
When he lands his horizontal velocity is 
<span>28 cos40 = 21.45 m/s </span>

<span>the time in flight comes from </span>
<span>x = Vht </span>
<span>58.8 = 21.45t </span>
<span>t = 2.74 seconds </span>

<span>his vertical velocity at landing is </span>
<span>28 sin40 = -18 m/s </span>

<span>his vertical velocity equation is </span>
<span>v = V0 - gt </span>
<span>-18 = V0 - 9.81(2.74) </span>
<span>V0 = -18 + 9.81(2.74) </span>
<span>V0 = 8.88 </span>

<span>his velocity magnitude was </span>
<span>v = (8.88^2 + 21.45^2)^½ </span>
<span>v = 23.2 m/s ANSWER </span>

<span>his initial direction was </span>
<span>tanθ = 8.88/21.45 </span>
<span>θ = 22.5 degrees above the horizontal ANSWER </span>

<span>to find the time to the flight apex from launch </span>
<span>v = gt </span>
<span>8.88 = 9.81t </span>
<span>t = 0.905 s </span>

<span>in 0.905 s Eddie has risen how far above the edge </span>
<span>y = ½(9.81)(0.905^2) </span>
<span>y = 4 m </span>

<span>the remainder of the flight is all drop and takes 2.74 - 0.905 = 1.85 seconds </span>

<span>in 1.85 seconds he drops </span>
<span>y = ½(9.81)(1.85^2) </span>
<span>y = 16.7 m </span>

<span>so the height from the edge to the landing point is </span>
<span>16.7 - 4 = 12.7 m ANSWER</span>
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