In the aerials competition in skiing, the competitors 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. The end of a launch ramp is directed 63° above the horizontal. With this launch angle, a skier attains a height of 14.8 m above the end of the ramp. What is the skier’s launch speed?

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Arclight Arclight · Answer 1 · 2019-12-27T15:31:14+01:00

Answer:

Launch speed is equal to [tex]19.11[/tex] meter/second

Explanation:

Using the newton's equation, we know that

[tex]V^2 = U^ 2 + 2 *a*S[/tex]

Where V is the final speed , U is the initial speed, a is the acceleration and S is the distance

At the highest point of the trajectory the object;s speed reduces to zero. hence

[tex]V = 0[/tex]

While initial velocity is equal to [tex]U sin 63[/tex]°

[tex]U^2 = -2* a* S\\U = \sqrt{-2* a* S} \\[/tex]

Substituting the given values we get -

[tex]U_0 * Sin 63 = \sqrt{- 2* -9.8* 14.8} \\U_0 = \frac{sqrt{- 2* -9.8* 14.8}}{Sin 63} \\U _0 = \frac{17.031}{0.891} \\U_0 = 19.11[/tex]

Thus, launch speed is equal to [tex]19.11[/tex] meter/second