Solved problem: Projectile in slanting tunnel

In summary: The red dot is where the point of closest approach is on the ceiling. In my plot the tunnel slopes down as it goes to the right.In summary, the conversation discusses the problem of finding the initial speed and landing distance of a projectile in a tunnel with a sloping ceiling. The relevant equations are derived and simplified, taking into account various angles and parameters. It is determined that the maximum distance is achieved when the projectile is parallel to the ceiling, and that the landing distance does not depend on the acceleration due to gravity. An alternate method for solving the problem is also presented, using geometric principles. The conversation also touches on the importance of correctly defining angles and the sign convention used for them. Finally, a plot is provided to visualize
  • #1
kuruman
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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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  • #2
kuruman said:
It is interesting to note that the landing distance d does not depend on g.
… which is as implied by dimensional analysis.
 
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  • #3
haruspex said:
… which is as implied by dimensional analysis.
Yes. The parabolic path ##y(x)=Ax^2+Bx+C## has three independent parameters and three conditions

##y(0)=0~;~~\left. \dfrac{dy}{dx}\right | _{x=0}\!=\tan\!\theta~;~~\left. \dfrac{dy}{dx}\right | _{x=x_p}\!=\tan\!\alpha.##

These determine the three parameters, ##C=0##, ##B =\tan\!\theta## and ##A=\dfrac{\tan\!\alpha-\tan\!\theta}{2x_p}##. Parameter ##x_p## scales as the dimension parameter ##L## of the tunnel and can be found by requiring that the point of closest approach lie on the ceiling.

In fact this presents an alternate method. Solve the geometric problem first to get an equation for the parabola, then write the parabolic trajectory ##y(x)## in terms of ##v_0## and ##g## and identify terms.
 
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  • #4
Am I understanding the setup? Here's how I see it for a specific case:

1684185875332.png


The blue line is the tunnel floor, orange is the tunnel ceiling. The trajectory is green.

For ##L = 5.0## m, ##\theta = 30^o##, and ##\alpha = 15^o##, the maximum range along the tunnel floor is ##d \approx 25## m for an initial speed ##v_0 \approx 14## m/s.

I find the general formulas corresponding to maximum range to be $$v_0 = \frac{\sqrt{2gL\cos \alpha}}{\sin(\alpha + \theta)}$$ $$d = 4L\left[\tan \alpha + \frac 1 {\tan(\alpha + \theta)}\right]$$

kuruman said:
$$d=\frac{\Delta x_{\!f}}{\cos\!\alpha}=\frac{4L}{\cos^2\!\alpha(\tan\!\theta-\tan\!\alpha)}.$$

This formula gives ##d = \infty## if ##\theta = \alpha##, which doesn't seem right. That's why I'm asking if I'm understanding the setup of the problem.
 
  • #5
You understand the setup correctly. However setting ##\theta =\alpha## is not a very good idea because that means that the projectile is shot at the surface and parallel to it. Your expressions and mine are identical when ##\alpha=0## but that doesn't mean much.

The argument ##(\alpha+\theta)## in the denominator of your expressions bothers me. In my expressions the angles are positive above the horizontal and negative below. That gets the sign of ##\Delta y## correctly. So in your example, I would enter a negative value for ##\alpha## because the tunnel is sloping down.

It is entirely possible that I made a mistake somewhere. I have to recheck my expressions.
 
  • #6
kuruman said:
The argument ##(\alpha+\theta)## in the denominator of your expressions bothers me. In my expressions the angles are positive above the horizontal and negative below. That gets the sign of ##\Delta y## correctly. So in your example, I would enter a negative value for ##\alpha## because the tunnel is sloping down.
OK, that clears it up! I used the opposite sign convention for ##\alpha##. Our formulas for the initial speed and range agree when the different sign conventions are taken into account. Thanks.
 
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  • #7
TSny said:
OK, that clears it up! I used the opposite sign convention for ##\alpha##. Our formulas for the initial speed and range agree when the different sign conventions are taken into account. Thanks.
All is well. Here is my plot reproducing yours.

ProjectileInTunnel.png
 
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