Proving tan\theta=(1-\sqrt{2})cot\theta

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Homework Help Overview

The problem involves a projectile launched from the midpoint between two inclined planes, both inclined at an angle \(\alpha\). The projectile's angle of projection is \(\alpha + \theta\), and it is required to prove the relationship \( \tan \theta = (1 - \sqrt{2}) \cot \theta \).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss rotating the axes to simplify the problem and consider the components of gravitational acceleration. There is mention of treating the situation as a standard kinematics problem. Some participants express confusion about the question's setup and the conditions under which the projectile touches both inclines.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the assumptions made. Some guidance has been offered regarding the treatment of the problem as a kinematics issue, but no consensus has been reached on the correct approach or the necessity of including \(\alpha\) in the proof.

Contextual Notes

There is uncertainty regarding the initial conditions and the relationship that needs to be proven, as well as the clarity of the projectile's trajectory in relation to the inclined planes.

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Homework Statement



There are two inclined planes each having an inclination of [tex]\alpha[/tex]. A projectile is launched from the midpoint between these two inclines such that the distance from the projectile on the ground to the inclines is x on either side.The angle of projection is [tex]\alpha[/tex] + [tex]\theta[/tex]and the initial velocity is u. The projectile just grazes the lower incline(having acute angle or [tex]\alpha[/tex] towards it) and lands on the other incline with an angle of 90 degrees to the incline.

Homework Equations



how do we prove that :

tan [tex]\theta[/tex]=(1-[tex]\sqrt{2}[/tex])cot [tex]\theta[/tex]

The Attempt at a Solution



Rotate the axis by [tex]\alpha[/tex] along the incline at the midpoint so that we have the angle of projection as [tex]\theta[/tex].The component acting downwards instead of g would be gcos [tex]\theta[/tex] and backwards gsin [tex]\theta[/tex].Next we equate it.Right?
 
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ron_jay said:

Homework Statement



There are two inclined planes each having an inclination of [tex]\alpha[/tex]. A projectile is launched from the midpoint between these two inclines such that the distance from the projectile on the ground to the inclines is x on either side.The angle of projection is [tex]\alpha[/tex] + [tex]\theta[/tex]and the initial velocity is u. The projectile just grazes the lower incline(having acute angle or [tex]\alpha[/tex] towards it) and lands on the other incline with an angle of 90 degrees to the incline.

Homework Equations



how do we prove that :

tan [tex]\theta[/tex]=(1-[tex]\sqrt{2}[/tex])cot [tex]\theta[/tex]

The Attempt at a Solution



Rotate the axis by [tex]\alpha[/tex] along the incline at the midpoint so that we have the angle of projection as [tex]\theta[/tex].The component acting downwards instead of g would be gcos [tex]\theta[/tex] and backwards gsin [tex]\theta[/tex].Next we equate it.Right?

Good idea to rotate the axes... I'm guessing there's supposed to be an [tex]\alpha[/tex] in that equation you need to prove. also the downward component of gravity would be [tex]gcos\alpha[/tex] and the backward would be [tex]gsin\alpha[/tex]

Not sure what you meant by equate it... but treat it like an ordinary kinematics problem now... see what you can get from the kinematics equations. Note that when the object lands, the backward component of velocity is 0 (along the new axes)...
 
i didn't get the question.probablya diagram should help.
well if the projectile is launcehd in between them how cum it touches both of them?
 
Good idea to rotate the axes... I'm guessing there's supposed to be an [tex]\alpha[/tex].

Not sure what you meant by equate it... but treat it like an ordinary kinematics problem now... see what you can get from the kinematics equations. Note that when the object lands, the backward component of velocity is 0 (along the new axes)...

i didn't get the question.probablya diagram should help.
well if the projectile is launcehd in between them how cum it touches both of them?

Here's a diagram which should clear your understanding.
 

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ron_jay said:
Here's a diagram which should clear your understanding.

Are you sure that's what you're supposed to prove? I went through the problem and I get that relationship but I get [tex]cot\alpha[/tex] on the right side.
 

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