Question

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?

Answer 1

Answer:

 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 $m/s^2$ 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 , $s= ut+\frac{1}{2} at^2$, 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 $m/s^2$ and time taken = 2u sin θ /g  

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$  

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$, 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

$0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}$

We have θ = 63° and H = 11.8 meter.

Substituting

   $11.8=\frac{u^2sin^263}{2*9.81}\\ \\ u^2=\frac{11.8*2*9.81}{sin^263}=\frac{231.516}{0.794} =291.58\\ \\ u=17.08m/s$

 The skier’s launch speed = 17.08 m/s

Answer 2

Answer:

Answer is 15.8 m/s.

Explanation:

A skier is moving down an incline and propelled at an edge 63 degrees over the even.  

Give v a chance to be the dispatch speed of the skier.  

The segment of speed along the flat bearing is vx = vcos÷θ

and the vertical part is vy=vsin÷θ

As the skier achieves a greatest stature after propelled is 10.1 m.  

Apply kinematic conditions to ascertain the skier speed in a vertical heading.  

vf²=vy²+2gh  

Here vf is the last speed, vy is the underlying speed in the vertical heading.  

At most extreme stature, last speed is zero.  

Thus,the intial vertical veloicty is,  

0²=vy²+2(- g)h  

v0 = √{2gh} =14.077 m/s  

In this way, the speed of the skier is,  

v0=vsin63  

v=v0÷sin63

=14.077m/s ÷ sin63 = 15.79 m/s  

In this way, the dispatch speed of the skier is 15.8 m/s