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 11.8 m above the end of the ramp. what is the skier’s launch speed?

 The skier’s launch speed = 17.08 m/s

Explanation:

The motion of skiing is projectile motion.

 Projectile motion has two types of motion Horizontal and Vertical motion.  

Vertical motion:  

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.  

Considering upward vertical motion of projectile.  

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g and final velocity = 0 m/s.  

0 = u sin θ - gt  

t = u sin θ/g  

Total time for vertical motion is two times time taken for upward vertical motion of projectile.  

So total travel time of projectile = 2u sin θ/g  

Horizontal motion:  

We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.  

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 and time taken = 2u sin θ /g  

So range of projectile,  

Vertical motion (Maximum height reached, H) :

We have equation of motion, , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.  

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

We have θ = 63° and H = 11.8 meter.

Substituting

 The skier’s launch speed = 17.08 m/s

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