<u>Answer:</u>
The skier’s launch speed = 17.08 m/s
<u>Explanation:</u>
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
![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](https://tex.z-dn.net/?f=11.8%3D%5Cfrac%7Bu%5E2sin%5E263%7D%7B2%2A9.81%7D%5C%5C%20%5C%5C%20u%5E2%3D%5Cfrac%7B11.8%2A2%2A9.81%7D%7Bsin%5E263%7D%3D%5Cfrac%7B231.516%7D%7B0.794%7D%20%3D291.58%5C%5C%20%5C%5C%20u%3D17.08m%2Fs)
The skier’s launch speed = 17.08 m/s