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(A) Let the velocity of projection be $v = 2\sqrt{gh}$ at an angle $	heta$ with the horizontal.Horizontal component of velocity: $v_x = v \cos 	heta

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$\sqrt{\frac{4h}{g}}$

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(A) Let the velocity of projection be $v = 2\sqrt{gh}$ at an angle $	heta$ with the horizontal.Horizontal component of velocity: $v_x = v \cos 	heta = 2\sqrt{gh} \cos 	heta$.The time taken to cover the horizontal distance $d = 2h$ between the walls is $t = \frac{d}{v_x} = \frac{2h}{2\sqrt{gh} \cos 	heta} = \sqrt{\frac{h}{g}} \sec 	heta$ ... $(i)$Using the equation of trajectory $y = x 	an 	heta - \frac{gx^2}{2v^2 \cos^2 	heta}$,for the two walls at $x_1$ and $x_2$ where $y = h$:$h = x 	an 	heta - \frac{gx^2}{2(4gh) \cos^2 	heta} \Rightarrow h = x 	an 	heta - \frac{x^2}{8h \cos^2 	heta}$.Rearranging gives $x^2 - (8h 	an 	heta \cos^2 	heta) x + 8h^2 = 0$.The distance between the walls is $x_2 - x_1 = 2h$. For a quadratic $ax^2 + bx + c = 0$,$|x_2 - x_1| = \frac{\sqrt{D}}{|a|}$.$|x_2 - x_1| = \frac{\sqrt{(8h 	an 	heta \cos^2 	heta)^2 - 4(8h^2)}}{1} = 2h$.$64h^2 	an^2 	heta \cos^4 	heta - 32h^2 = 4h^2 \Rightarrow 16 \sin^2 	heta \cos^2 	heta - 8 = 1 \Rightarrow 4 \sin^2(2	heta) = 9$ (This implies a specific geometry).Alternatively,using the time of flight between walls $t = \frac{2v_y}{g}$,where $v_y$ is the vertical velocity at height $h$: $v_y^2 = (v \sin 	heta)^2 - 2gh = 4gh \sin^2 	heta - 2gh = 2gh(2\sin^2 	heta - 1)$.$t = \frac{2\sqrt{2gh(2\sin^2 	heta - 1)}}{g} = 2\sqrt{\frac{2h}{g}(2\sin^2 	heta - 1)}$.Equating the two expressions for $t$: $\sqrt{\frac{h}{g}} \sec 	heta = 2\sqrt{\frac{2h}{g}(2\sin^2 	heta - 1)}$.$\sec^2 	heta = 8(2\sin^2 	heta - 1) = 16\sin^2 	heta - 8$.$1 + 	an^2 	heta = 16\sin^2 	heta - 8 \Rightarrow 	an^2 	heta = 3 \Rightarrow 	heta = 60^\circ$.Substituting $	heta = 60^\circ$ into $(i)$: $t = \sqrt{\frac{h}{g}} \sec 60^\circ = 2\sqrt{\frac{h}{g}} = \sqrt{\frac{4h}{g}}$.

$A$ particle is projected with velocity $2 \sqrt{gh}$,so that it just flies over two walls of equal height $h$ and $2h$ distance apart from each other. Find the time for which the particle flies between the walls.

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