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

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The discussion revolves around proving the equation tan θ = (1 - √2) cot θ in the context of a projectile launched between two inclined planes. Participants suggest rotating the axes to simplify the problem and clarify the angle of projection. There is confusion regarding the components of gravity, with some suggesting they should be expressed in terms of α rather than θ. A diagram is recommended to enhance understanding of the projectile's trajectory and its interaction with the inclines. Overall, the conversation emphasizes the need for clarity in the problem setup and the correct application of kinematic equations.
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Homework Statement



There are two inclined planes each having an inclination of \alpha. 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 \alpha + \thetaand the initial velocity is u. The projectile just grazes the lower incline(having acute angle or \alpha 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 \theta=(1-\sqrt{2})cot \theta

The Attempt at a Solution



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

Homework Statement



There are two inclined planes each having an inclination of \alpha. 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 \alpha + \thetaand the initial velocity is u. The projectile just grazes the lower incline(having acute angle or \alpha 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 \theta=(1-\sqrt{2})cot \theta

The Attempt at a Solution



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

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

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 \alpha.

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 cot\alpha on the right side.
 
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