Question

In the What If? Section of Example 4.5 it was claimed that the maximum range of a ski jumper occurs for a launch angle$\mathit{\theta }$ given by

role="math" localid="1663692909020" $\mathit{\theta }\mathbf{=}{\mathbf{45}}^{\mathbf{\circ }}\mathbf{-}\frac{\mathbf{\varphi }}{\mathbf{2}}$

Where$\mathit{\varphi }$ is the angle the hill makes with the horizontal in Figure 4.14. Prove this claim by deriving the equation above.

Short answer

The maximum range of a ski jumper occurs for a launch angle $\theta$ given by $\theta =45°-\frac{\varphi }{2}$ is proved.

Step-by-step solution

Definition of Range of Projectile.

The horizontal displacement of the projectile determines its maximum or minimum range. Gravity most effective acts vertically, consequently there may be no acceleration on this direction indicates the variety line.

The variety of the projectile, like its time of flight and most height, is a feature of its preliminary speed.

To prove the given θ  value by using projection method.

If the point of projection is taken as the origin, and if the point of landing has coordinates$\left(x_0-y\right)$then, using the equation for the trajectory, we have

$\begin{array}{c}-y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }\\ -\frac{y}{x}=\tan \theta -\frac{gx}{2u^2{\cos }^2\theta }\end{array}$

Using $\frac{y}{x}=\tan \varphi ,$

$\begin{array}{rcl}-\tan \varphi & =& \tan \theta -\frac{gx}{2u^2{\cos }^2\theta }\\ x& =& \frac{2u^2{\cos }^2\theta }{g}\left(\tan \theta +\tan \varphi \right)\end{array}$

Determine the range of projectile.

The range of the projectile on the incline is

$\begin{array}{c}R=x\sec \varphi \\ R=\frac{2u^2{\cos }^2\theta }{g}\left(\tan \theta +\tan \varphi \right)\times \sec \varphi \\ R\left(\theta \right)=\frac{2u^2\sin (\theta +\varphi )\cos \theta }{g{\cos }^2\varphi }\end{array}$

The range is a function of $\theta$. To maximize it,

We set $\frac{\partial R\left(\theta \right)}{\partial \theta }=0$

$\begin{array}{c}\frac{\partial }{\partial \theta }\left[\frac{2u^2\sin (\theta +\varphi )\cos \theta }{g{\cos }^2\varphi }\right]=0\\ \frac{\partial }{\partial \theta }\left[\frac{u^2}{g{\cos }^2\varphi }\left(\sin (2\theta +\varphi )+\sin \varphi \right)\right]=0\\ 2\cos (2\theta +\varphi )=0\end{array}$

$\cos (2\theta +\varphi )$ is a maximum when$2\theta +\varphi =\frac{\pi }{2}$

$\begin{array}{c}\theta =\frac{\pi }{4}-\frac{\varphi }{2}\\ \theta =45°-\frac{\varphi }{2}\end{array}$

Hence, the maximum range of a ski jumper occurs for a launch angle $\theta$given by $\theta =45°-\frac{\varphi }{2}$ is proved.