Building a ski jump

Whoa this problem was fun! Ok, so here's watcha do (at least what I did!):

First, we draw a picture of the situation (you, the reader, should do this!). We have an initial velocity $v$ for this skier, and it makes an angle $\theta$ with the horizontal, so let's split the velocity into two components: the horizontal, or x component $v\cos\theta$ and the vertical, or y component $v\sin\theta$ . Now we make the equations of motion for the two dimensions:

$v\sin\theta t-\frac{1}{2}gt^2=-y$

$v\cos\theta t=x$

Now we find t as a function of x: $t=\frac{x}{v\cos\theta}$ and replace this solution into our equation for the y dimension (I skip the basic algebra):

$\tan\theta x-\frac{\frac{1}{2}gx^2}{(v\cos\theta)^2}=-y$

If you drew a picture (smart thing to do!), you'll see that we can make a simple relationship between y and x, since $\tan20=\frac{y}{x}$ . Mainly, that $y=x\tan20$ , so we put this into our last equation:

$\tan\theta x-\frac{\frac{1}{2}gx^2}{(v\cos\theta)^2}=-x\tan20$

And doing some simple algebra, we find that

$x=\frac{(\tan\theta+\tan20)2(v\cos\theta)^2}{g}$

Up next, we take the derivative of x with respect to $\theta$ and equal it to zero. We do this because we want to see for which angle $\theta$ , x is at its maximum (for which angle does the skier go farthest?). Pretty simple derivatives for the most part, so I leave that to you. As a final result, we have that

$0=1-2\sin^2\theta-2\sin\theta\cos\theta\tan20$

Now we'll need some trigonometric identities voodoo to fix this messy-looking thing. Remember that $\sin^2\theta=\frac{1}{2}(1-\cos2\theta)$ ? No? Well it's true! Also, a more well-known identity: $2\sin\theta\cos\theta=\sin2\theta$ . Now things are looking much more interesting! We replace some stuff, and arrive at

$\tan2\theta=\frac{1}{\tan20}$

$2\theta=70$

$\theta=35$

Whoooo!

According to the question, the diagram is as follows: IMAGE

We rotate the figure $20^o$ anticlockwise. In this figure, only the portion where ski jump takes place is shown. Here, $\alpha = \theta +20^o$ IMAGE

According to the above figure,

Total change in height=0. Let the time of flight be $t$ and the velocity of the skier be $u$ .

$\therefore 0=usin \alpha t- \frac{1}{2} gcos20^o t^2$ $\implies t= \frac{2usin \alpha}{gcos20^o}$ (t can't be 0).

Range= $ucos \alpha t + \frac{1}{2}gsin20^ot^2$

$\implies Range= \frac{2u^2sin \alpha}{gcos^220^o}(cos \alpha cos20^o+sin \alpha sin20^o)$

$\implies Range=\frac{2u^2}{gcos^220^o}sin \alpha cos( \alpha -20^o)$

Hence, range will be maximum if $sin \alpha cos( \alpha -20^o)$ is maximum.

Differentiating the above expression w.r.t $\alpha$ we get:

$cos \alpha cos(\alpha -20^o)-sin \alpha sin(\alpha -20^o)=cos(2 \alpha -20^o)$

For maximum value, the above expression should be equal to 0 which gives $\alpha =55^o$ . Since $\alpha = \theta +20^o$ , we get $\theta =35^o$ .

Let $d$ be the distance travelled on the ramp, and $\phi$ be the angle of incline. Firstly, if $\theta=0$ , then then $d$ will be shorter. if theta increases, $d$ increases, but up to a certain point. After that (for instance if $\theta = 90$ , then $d$ will be 0. So we are trying to find the maximum of $d$ using differentiation.

$x_f = v_i\cos \theta \cdot t = d\cos \phi$ $[1]$

$y_f = v_i\sin \theta \cdot t - \frac{1}{2}gt^{2} = -d\sin \phi$ $[2]$

From $[1]$ : $t = \frac{d\cos \phi}{v_i\cos \theta}$ $[3]$

Substituting $[3]$ into $[2]$ :

$v_i\sin \theta \cdot (\frac{d\cos \phi}{v_i\cos \theta}) - \frac{1}{2}g(\frac{d\cos \phi}{v_i\cos \theta})^{2} = -d\sin \phi$

$d\cos\phi \tan\theta - \frac{gd^{2}\cos^{2}\phi}{2v_i^{2}\cos^{2}\theta} = -d\sin\phi$

It will be useful to express the equation in terms of $d$ , since we want to find the maximum of $d$ so we can proceed to divide the whole equation by $d$ .

$\cos\phi\tan\theta - \frac{gd\cos^{2}\phi}{2v_i^{2}\cos^{2}\theta} = -\sin\phi$

Now we will express the equation in terms of $d$ .

$d = \frac{2v_i^{2}\cos^{2}\theta(\cos\phi\tan\theta+\sin\phi)}{gcos^{2}\phi}$

Simplifying it we get

$d = \frac{v_i^{2}\sin 2\theta}{g\cos\phi} + \frac{2v_i^{2}\cos^{2}\theta\tan\phi}{g\cos\phi}$

In order to find maximum $\theta$ , we have to differentiate, keeping in mind that $\phi$ is constant, and the only variables are $d$ and $\theta$ .

$\frac{\mathrm {d}d}{\mathrm{d}\theta}=\frac{v_i^{2}}{g\cos\phi}\cdot2\cos 2\theta - \frac{2v_i^{2}\tan\phi}{g\cos\phi}\cdot\sin 2\theta$

Setting $\frac{\mathrm {d}d}{\mathrm{d}\theta}=0$ , to find our maximum, we get

$\frac{v_i^{2}}{g\cos\phi}\cdot2\cos 2\theta - \frac{2v_i^{2}\tan\phi}{g\cos\phi}\cdot\sin 2\theta=0$

$2v_i^{2}\cos 2\theta - 2v_i^{2}\tan\phi\sin 2\theta =0$

$\cos 2\theta-\tan\phi\sin 2\theta=0$

Using the R-Formulae trigonometric identity

$\sqrt{\tan^{2}\phi+1}\cos(2\theta+\phi) = 0$

$cos(2\theta+\phi) = 0$

$2\theta + \phi = 90^{\circ}$

$\theta = 45^{\circ} - \frac{\phi}{2}$

Substituting $\phi = 20^{\circ}$ , we get $\theta = 35^{\circ}$

Consider the trajectory in the Cartesian plane. Let the ski jump be at the point $(0,0)$ . Then the trajectory of the skier can be described as

$$y = -\frac{g}{2v^2 \cos{\theta}} x^2 + x\tan{\theta}$$

and the hill as

$$y = - x \tan{\alpha}$$.

Equating $y$ 's to find the $x$ -coordinate of the landing point (and discarding the trivial case $x=0$ ) we get

$$x = \frac{(\tan{\theta} + \tan{\alpha}) 2v^2 \cos^2{\alpha}}{g}$$.

The distance down the slope $d$ in this case is just $d = x /\cos{\alpha}$ .

To find $\theta$ , maximise $d$ w.r.t. $\theta$ (which is a lengthy and quite boring derivation), from where it follows that $\theta \approx 0.61 \text{rad} \approx \boxed{35 ^\circ}$ .