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Two parallel straight lines are inclined to the horizontal at an angle $α$ . A particle is projected from a point mid way between them so as to graze one of the lines and strike the other normally. If $θ$ is the angle between the direction of projection and either of lines, then $\tan θ = (\sqrt{p} - 1) \cot α$ then p is

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answer is 2.

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$v_{y}^{2} = u_{y}^{2} + 2 a_{y} y \Rightarrow 0 = {(u \sin θ)}^{2} - 2 (g \cos α) b \Rightarrow b = \frac{u^{2} \sin^{2} θ}{2 g \cos α} ...... (1)$

Using $v_{y} = u_{y} + a_{x} t$
$0 = u \cos θ - (g \sin α) t$
$t = \frac{u \cos θ}{g \sin α} ...... (2)$
For motion along y-direction we have
$y = u_{y} t + \frac{1}{2} a_{y} t^{2}$
$\Rightarrow - b = u \sin θ (\frac{u \cos θ}{g \sin α}) + \frac{1}{2} (- g \cos α) {(\frac{u \cos θ}{g \sin α})}^{2}$

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