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vlabodo [156]
3 years ago
12

The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply

upward at the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. In the women's competition, the end of a typical launch ramp is directed 61° above the horizontal. With this launch angle, a skier attains a height of 16 m above the end of the ramp. What is the skier's launch speed?

Physics
1 answer:
kykrilka [37]3 years ago
4 0

Answer: Got It!

<em>Explanation:  </em>let s = speed at launch

v = 0 at top = s sin 63 - g t

so at top

t = s sin 63/g = .0909 s

h = 13.6 = s sin 63 t - 4.9 t^2

13.6 = .081s^2 - .0405 s^2

s^2 = 336

s = 18.3 m/s

0  0

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B) H = 24.08 ft

C) M.A = 12.04

D) P = 13.7 lb

Explanation:

A)

Minimum allowable length of base of ramp can be found as follows:

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where,

Slope = 1/12

H = Height of Ramp = 2 ft

B = Length of Base of Ramp = ?

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B = 2 ft * 12

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B)

The length of the slope of ramp can be found by using pythagora's theorem:

L = √H² + B²

where,

H = Perpendicular = height = 2 ft

B = Base = Length of Base of Ramp = 24 ft

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Therefore,

H = √[(2 ft)² + (24 ft)²]

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The mechanical advantage of an inclined plane is given by the following formula:

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Another general formula for Mechanical Advantage is:

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