# Projectile Prob

1. Aug 18, 2007

### ron_jay

1. The problem statement, all variables and given/known data

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 $$\alpha$$ + $$\theta$$and 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.

2. Relevant equations

how do we prove that :

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

3. 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?

2. Aug 18, 2007

### learningphysics

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)...

3. Aug 18, 2007

### pardesi

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?

4. Aug 19, 2007

### ron_jay

Here's a diagram which should clear your understanding.

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5. Aug 19, 2007

### learningphysics

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.