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A skilled skier knows to jump upward before reaching a downward slope. Consider a jump in which the launch speed is v0=10m/s, the launch angle is θ=11.3° , the initial course is approximately flat, and the steeper track has a slope of 9.0° . Figure 4-42a shows a prejump that allows the skier to land on the top portion of the steeper track. Figure 4-42b shows a jump at the edge of the steeper track. In Fig. 4-42a, the skier lands at approximately the launch level. (a)In the landing, what is the angle φ between the skier’s path and the slope? In Fig. 4-42b, (b) how far below the launch level does the skier land, and (c) what is ? (The greater fall and greater φcan result in loss of control in the landing.)

Short Answer

Expert verified
  1. The angle between the skier’s path and the slope is 2.3°
  2. Skier land approximately 1.1 m below the launch level.
  3. The angle φis18° .

Step by step solution

01

Given data

  1. Initial speed of launch is v0=10m/s.
  2. Angle of launch is θ=11.3°.

3. The slope of the track isθ0=9°

02

Understanding the physical situation and use of kinematic equations

Knowing the slope angle, the initial velocity, and the angle with which the skier jumps and using the kinematic equations, the concept of projectile motion, it is possible to find out the angle betweenthe skier’s path and the slope, and the distance between the initial position and final position.To find the angle between slope and velocity vector, the skier’s jump angle with horizontal and the angle between snow surface and horizontal is sufficient.The first kinematic equation is useful to find the components of velocity at a time ‘t’, which is required to find the direction of travel. The second kinematic equation is used to find the distance between the skier and the launch level.

03

Formulae for kinematic equations and direction of travel at landing

First kinematic equation is,

v=v0+at …(i)

Second kinematic equation is,

role="math" localid="1661144647913" s=v0t+12at2 …(ii)

θ=tan-1VyVx … (iii)

04

(a) Calculation of the angle between the slope and the velocity vector

The skier jumps at an angleθ=11.3° . The skier will return to the horizontal level with the same vector component but in the downward direction. Thesnow surface makes an angle θ0=9°downwards with the horizontal surface. Therefore, the angle between the slope and velocity vector is the difference between these two angles.

α=θ-θ0=11.3°-9.0°=2.3°

Therefore, the angle between the slope and velocity vector is 2.3°.

05

Step 4:(b) Calculation for the approximate distance between skier and launch level

Now, let the skier land at a distance d down the slope.

So,

x=dcosθ0&y=-dsinθ0

We have from equation (ii)

-dsinθ0=-dcosθ0tanθ-gdcosθ02v02cos2θ

Solving for d, we obtain

role="math" localid="1661145150417" -d=2v02cos2θgcosθ02cosθ0.tanθ+sinθ0=2v02cosθgcos2θ0cosθ0.sinθ+cosθ.sinθ0=2v02cosθgcos2θ0sinθ+θ0

If we substitute all the values we can find,

d=210m/s2cos11.3°9.8m/s2cos2(9°)sin(11.3°+9°)=7.117m

This gives

y=-dsinθ0=-7.117msin(9°)=-1.1m

So, skiers land approximately 1.1 m below the launch level.

06

 Step 5: Calculation of time taken for skier to land

We know,

t=xvx=7.27mcos9°10m/scos9°=0.72s

07

(c) Determining the direction of travel

18°We can calculate velocity components

vx=v0cosθ=10m/scos9°=9.87m/s

vy=sinθ-gt=(10m/s)sin(9.0)-(9.8mls2)(0.72s)=5.55m/s

θ=tan-1vyvx=tan-1-5.5m/s9.87m/s=-29.1°

Thus, the direction of travel is 29.1 degree below the horizontal.

Hence,

ϕ=29.1-11.3=17.8o180

Therefore, the angle between the skier’s path and the slope is 9o.

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