Ski jumping

Note: ski jump is parallel to the hill, and therefore angled at 30°

First, let us find the velocity of the skier upon "launching". Since there is some friction between skier and ice, we have:

$\Delta E = A = - F_{fr} \cdot L$

where $L$ is length of the ski jump, which is $L = H / sin(\theta)$ , $H=50m$ and $F_{fr} = \mu \cdot mg \cdot cos(\theta)$

Also, $\Delta E = E_{2} - E_{1} = \frac {1}{2}mv^2 - mgH$ , so for the velocity we have:

$v^2 = 2gH - 2gH \cdot \mu / tg(\theta)$ , which gives: $v=29.919 m/s$

Now, the fun part. We should choose our coordinate system to be placed at the launch point, or at the impact point. Either will work in this case, and I will chose the impact point, just to demonstrate how it is done.

In this system, initial position is $(xo,yo)$ , and initial velocity is $(-vo,0)$ - only in x-direction, and the final position is $(0,0)$ . Motions in $x$ and $y$ directions are mutually independent (since there is no air friction), so we have:

$x = x_{o} - v_{o} t \$ and $\ y = y_{o} - \frac {1}{2} gt^2$

and at the moment of impact, $(x,y)=(0,0)$ , so:

$x_{o} = v_{o} t \$ and $\ y_{o} = \frac {1}{2} gt^2$

If you draw the impact point on the figure in the original figure, you should be able to see that: $y_{o} = x_{o} \cdot tg(\theta) + h\$ , where $h=5m$

Using the last three equations we get the following quadratic equation:

$\frac{1}{2}g \cdot t^2 - v_{o}\ tg(\theta) \cdot t - h = 0$

which gives two solutions:

$t_{1} = -0.269 sec$

$t_{2} = 3.794 sec$

First result would have had physical sense if the skier had been in flight even before the launch, which we know not to be true, so the final solution is: $t = 3.794 sec$

\noindent Let us split this scenario into two parts and analyse them separately to simplify matters:\

\begin{enumerate}

\item \textbf{Skiing on the slope} \item \textbf{Flying in mid-air (after launch)}\[5pt] \end{enumerate}

\noindent \mbox{\bfseries 1. Skiing on the slope}\[7pt] Before jumping straight into this part, let us look at part 2 briefly. Since we want to find the time t, it is logical that we should find the speed of launch (which we shall denote by $v$). The common and widely-used method to find $v$ is by conservation of energy.\[5pt] The initial energy that the skier has is his or her gravitational potential energy, $Mgh$. The final forms of energy the skier has (when he or she reaches the launching point) should be his or her kinetic energy, $\frac{1}{2}Mv^2$.\[5pt] However, the matter is not so simple. Since we are given the coefficient of friction $\mu$, we have to take the work done by the frictional force, $f$, into account. Thus, we get the equation below,

\begin{equation} Mgh=\frac{1}{2}Mv^2+fd \end{equation}

\noindent where $d$ is the distance (NOT displacement, since friction is a non-conservative force) travelled by the skier. We then note that the only unknown terms in the equation above are $f$ and $d$.\[5pt] To find $f$, we use the relation $f=\mu N$, where $N$ is the normal force exerted by the ski slope on the skier. We then draw the free-body diagram to find $N$(since I haven't learnt how to draw diagrams in \LaTeX , I shall describe it in words).\[5pt] The forces acting on the skier are friction, gravity (weight), and the normal force. Let us define the x and y planes as the planes parallel to and perpendicular to the surface of the ski slope respectively. Since $N$ is in the y-plane, we only need to care about the components of the other forces in the y-plane, which in this case, can only be that of gravity. Since there is no net acceleration in the y-plane, the net sum of the forces in the y-plane is zero. This means $N=Mg\cos 30^{\circ}$.\[5pt] As for $d$, we realise that it is just the length of the ski slope. Hence, $d=h\csc 30^{\circ}$.\[5pt] Now that the only unknown term in equation (1) is $v$, we can find $v$.\[7pt] \noindent \mbox{\bfseries 2. Flying in mid-air (after launch)}\[7pt] It should be clear that in this part, our kinematic equations come into play.\[5pt] In the vertical direction, we have \begin{equation} \frac{1}{2}gt^2=5+m \end{equation} where $m$ is the vertical displacement of fall between the launch point and the point on the slope the skier impacts.\[5pt] In the horizontal direction, we have \begin{equation} vt=n \end{equation}
where $n$ is the horizontal displacement of fall between the launch point and the point on the slope the skier impacts.\[5pt] Next, we realise that there is a trigonometric relation between $m$ and $n$. It is \begin{equation} m=n\tan 30^{\circ} \end{equation} After that, we substitute equation (4) into (2), then equation (2) into equation (3). We will get a quadratic equation with $t$ as the only unknown variable. It should be as follows: \begin{equation} \frac{\sqrt{3}}{2}gt^2-\sqrt{\left(100+5\sqrt{3}\right) g}\hspace{1mm}t-5\sqrt{3}=0 \end{equation} Solving equation (5) should give us the value of $t$.

The net force on the skier on the ski jump parallel to the ski jump surface is a combination of the gravitation force and the frictional force,

\(F_{net}=mg(sin \theta - \mu cos\theta)\).

The net work done on the skier as he travels down the jump is $W=FL$ where L is the length of the jump, $L=50/sin \theta$ . By the work-energy theorem the skier launches horizontally from the bottom of the jump with a speed

$v=(2gh(1-\mu cot \theta))^{1/2}$ .

After the skier launches (and treating the launch point as x,y=0), the horizontal and vertical position of the skier are

$x=vt$ , $y=-\frac{1}{2}gt^2$ .

The landing spot is on the line $y=-xtan \theta-5$ , so we have a system of equations we can solve for t=3.794 seconds.

Using the conservation of energy principle, the work done by friction must be equal to the change in the mechanical energy of the system (h = 50m, alpha = 30°),

$\frac{1}{2} m v^2 - m g h = - \mu m g \cos{\alpha} \frac{h}{\sin{\alpha}}$

which implies

$v = \sqrt{2 g h (1 - \mu cot \alpha)}$ .

During the fall, using parallel (x axis pointing 'downwards') and perpendicular (y axis pointing 'upwards') axes in respect to the hill, we can write down the variation of y with time as

$y = y_0 + v \sin{\alpha t} - \frac{g \cos{\alpha} t^2}{2}$ .

Setting y = 0 and solving for t (h' = 5m),

$t = \frac{v \tan{\alpha}}{g} \left( 1 + \sqrt{1 + \frac{2 g h'}{v^2 \tan{\alpha}^2}} \right) = 3.79 s$ .

The net force on the skier on the ski jump parallel to the ski jump surface is a combination of the gravitation force and the frictional force, . The net work done on the skier as he travels down the jump is where L is the length of the jump, . By the work-energy theorem the skier launches horizontally from the bottom of the jump with a speed . After the skier launches (and treating the launch point as x,y=0), the horizontal and vertical position of the skier are ,. The landing spot is on the line , so we have a system of equations we can solve for t=3.794 seconds.