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- Homework Statement
- A rock is projected at angle θ relative to the horizontal inside a square tunnel of side L that is slanting at angle α relative to the horizontal (α < θ). The rock starts at floor level and describes a trajectory that maximizes the landing distance. Ignore air resistance.

(a) Find the initial speed.

(b) Find the landing distance.

Inspired by https://www.physicsforums.com/threads/rock-falling-down-a-tunnel.1052414/#post-6888206

- Relevant Equations
- See below.

**(a) Find the initial speed.**

__Solution__

The relevant equations are (4) and (5) derived here and reproduced below as (1) and (2) respectively:

$$\begin{align}

& \frac{\Delta x}{\tan\!\theta-\tan\!\varphi}=\frac{2v_{0x}^2}{g} \\

& \tan\!\varphi=\frac{1}{2}(\tan\!\theta+\tan\!\omega).

\end{align}$$In these equations all angles are relative to the horizontal and defined as

##\theta =## angle of projection; ##\varphi=## angle of position vector; ##\omega=## angle of velocity vector.

The other symbols have their usual meanings.

The distance traveled is maximized when the rock is at the ceiling and traveling parallel to it. That's the only trajectory that will get it as close as possible to the ceiling without hitting it. This means that ##\omega =\alpha.## Equation (2) becomes

$$\tan\!\varphi=\frac{1}{2}(\tan\!\theta+\tan\!\alpha)\implies \tan\!\theta-\tan\!\varphi=\frac{1}{2}(\tan\!\theta-\tan\!\alpha).$$Let ##\{\Delta x_p,\Delta y_p\}## be the point of closest approach to the ceiling. We get ##\Delta x_p## from Equation (1),

$$\frac{\Delta x_p}{\frac{1}{2}(\tan\!\theta-\tan\!\alpha)}=\frac{2v_{0x}^2}{g}\implies \Delta x_p=\frac{v_{0x}^2}{g}(\tan\!\theta-\tan\!\alpha)$$ and $$\Delta y_p=\Delta x_p\tan\!\varphi=\left[ \frac{v_{0x}^2}{g}(\tan\!\theta-\tan\!\alpha) \right]\times\frac{1}{2}(\tan\!\theta+\tan\!\alpha)=\frac{v_{0x}^2}{2g}(\tan^2\!\theta-\tan^2\!\alpha).$$Using simple trigonometry, we get the straight-line equation for the ceiling on which the point of closest approach must lie, $$

\begin{align}

& y= (\tan\!\alpha) ~x+\frac{L}{\cos\!\alpha} \nonumber \\

& \Delta y_p= (\tan\!\alpha) ~\Delta x_p+\frac{L}{\cos\!\alpha} \nonumber \\

& \frac{v_{0x}^2}{2g}(\tan^2\!\theta-\tan^2\!\alpha)=\tan\!\alpha \frac{v_{0x}^2}{g}(\tan\!\theta-\tan\!\alpha)+\frac{L}{\cos\!\alpha} \nonumber \\

& \frac{v_{0x}^2}{2g}(\tan\!\theta-\tan\!\alpha)^2=\frac{L}{\cos\!\alpha}\nonumber \\

& v_0=\frac{1}{\cos\!\theta} \left[\frac{2gL}{ \cos\!\alpha (\tan\!\theta-\tan\!\alpha)^2} \right]^{1/2}

\nonumber \end{align}$$

**(b) Find the landing distance.**

__Solution__

Here ##\tan\!\varphi=\tan\!\alpha## First we find the horizontal displacement using equation (1).

$$\begin{align}

& \frac{\Delta x_{\!f}}{\tan\!\theta-\tan\!\alpha}=\frac{2v_{0x}^2}{g} \nonumber \\

& \Delta x_{\!f}=\frac{4L}{\cos\!\alpha(\tan\!\theta-\tan\!\alpha)}. \nonumber \\

\end{align}$$ Then for the landing distance ##d##, $$d=\frac{\Delta x_{\!f}}{\cos\!\alpha}=\frac{4L}{\cos^2\!\alpha(\tan\!\theta-\tan\!\alpha)}.$$It is interesting to note that the landing distance ##d## does not depend on ##g##.

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