A skilled skier knows to jump upward before reaching a downward slope. Consider a jump in which the launch speed is ${\mathbf{v}}_{\mathbf{0}}\mathbf{=}\mathbf{10}\mathbf{m}\mathbf{/}\mathbf{s}$, the launch angle is $\mathbf{\theta }\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{3}\mathbf{°}$ , the initial course is approximately flat, and the steeper track has a slope of $\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{°}$ . 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 $\mathbf{\phi }$ 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 $\mathbf{\phi }$can result in loss of control in the landing.)

1. Initial speed of launch is ${\mathrm{v}}_0=10\mathrm{m}/\mathrm{s}$.
2. Angle of launch is $\theta =11.3°$.

3. The slope of the track is${\theta }_0=9°$

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.

First kinematic equation is,

$v=v_0+at$ …(i)

Second kinematic equation is,

role="math" localid="1661144647913" $s=v_{0}t+\frac{1}{2}a{t}^2$ …(ii)

$\theta ={\tan }^{-1}\left(\frac{V_y}{V_x}\right)$ … (iii)

The skier jumps at an angle$\theta =11.3°$ . The skier will return to the horizontal level with the same vector component but in the downward direction. Thesnow surface makes an angle ${\theta }_0=9°$downwards with the horizontal surface. Therefore, the angle between the slope and velocity vector is the difference between these two angles.

$$\begin{gathered} \alpha =\theta -{\theta }_0 \\ =11.3°-9.0° \\ =2.3° \end{gathered}$$

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

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

So,

$x=d\cos {\theta }_0\&y=-d\sin {\theta }_0$

We have from equation (ii)

$-d\sin {\theta }_0=-dcos{\theta }_0\tan \theta -\frac{g\left(d\cos {\theta }_0\right)}{2v_0^2{\cos }^2\theta }$

Solving for d, we obtain

role="math" localid="1661145150417" $$\begin{gathered} -\mathrm{d}=\frac{2{\mathrm{v}}_0^2{\cos }^2\mathrm{\theta }}{{\mathrm{gcos\theta }}_0^2}\left({\mathrm{cos\theta }}_0.\mathrm{tan\theta }+{\mathrm{sin\theta }}_0\right) \\ =2{\mathrm{v}}_0^2\frac{\cos \mathrm{\theta }}{\mathrm{g}{\cos }^2{\mathrm{\theta }}_0}\left({\mathrm{cos\theta }}_0.\mathrm{sin\theta }+\mathrm{cos\theta }.{\mathrm{sin\theta }}_0\right) \\ =2{\mathrm{v}}_0^2\frac{\cos \mathrm{\theta }}{\mathrm{g}{\cos }^2{\mathrm{\theta }}_0}\sin \left(\mathrm{\theta }+{\mathrm{\theta }}_0\right) \end{gathered}$$

If we substitute all the values we can find,

$$\begin{gathered} \mathrm{d}=\frac{2{\left(10\mathrm{m}/\mathrm{s}\right)}^2\cos \left(11.3°\right)}{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\cos }^2(9°)}\sin (11.3°+9°) \\ =7.117\mathrm{m} \end{gathered}$$

This gives

$$\begin{gathered} \mathrm{y}=-{\mathrm{dsin\theta }}_0 \\ =-\left(7.117\mathrm{m}\right)\sin (9°) \\ =-1.1\mathrm{m} \end{gathered}$$

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

We know,

$$\begin{gathered} \mathrm{t}=\frac{\mathrm{x}}{{\mathrm{v}}_{\mathrm{x}}} \\ =\frac{\left(7.27\mathrm{m}\right)\cos \left(9°\right)}{\left(10\mathrm{m}/\mathrm{s}\right)\cos \left(9°\right)} \\ =0.72\mathrm{s} \end{gathered}$$

$18°$We can calculate velocity components

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}={\mathrm{v}}_0\mathrm{cos\theta } \\ =\left(10\mathrm{m}/\mathrm{s}\right)\cos \left(9°\right) \\ =9.87\mathrm{m}/\mathrm{s} \end{gathered}$$

$\begin{array}{l}{\mathrm{v}}_{\mathrm{y}}=sin\mathrm{\theta }-\mathrm{gt}\\ =(10\mathrm{m}/\mathrm{s})sin(9.0)-(9.8{\mathrm{mls}}^2)(0.72\mathrm{s})\\ =5.55\mathrm{m}/\mathrm{s}\end{array}$

$$\begin{gathered} \mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{v}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(-\frac{5.5\mathrm{m}/\mathrm{s}}{9.87\mathrm{m}/\mathrm{s}}\right) \\ =-29.1° \end{gathered}$$

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

Hence,

$\begin{array}{l}\varphi =29.1-11.3\\ =17.8^o\\ \approx {18}^0\end{array}$

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