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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 10.9m above the end of the ramp. What is the skier’s launch speed?

Respuesta :

Answer:

u = 11.6 m/s

Explanation:

The end of a launch ramp is directed 63° above the horizontal. A skier attains a height of 10.9 m above the end of the ramp.

Maximum height, H = 10.9

Let v is the launch speed of the skier. The maximum height attained by the projectile is given by :

[tex]H=\dfrac{u^2\ sin^2\theta}{g}[/tex]

[tex]10.9=\dfrac{u^2\ sin^2(63)}{9.8}[/tex]

u = 11.6 m/s

So, the launch speed of the skier is 11.6 m/s. Hence, this is the required solution.

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