A particle is projected with velocity 2(√(gh)) so that it just clears two walls of equal heigh(h) in the t_1 and t_(2 ) respectively.The two walls are at a distance of 2h from each other.If time passing between the two walls is 2(√(h/g)) show that (i) angle projected 60^° (ii)t_1 +t


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Question Number 154549 by peter frank last updated on 19/Sep/21

$A particle is projected with velocity$ $2 \sqrt{gh} so that it just clears two$ $walls of equal heigh (h) in the$ $t_{1} and t_{2} respectively . The two$ $walls are at a distance of 2 h from$ $each other . If time passing$ $between the two walls is 2 \sqrt{\frac{h}{g}}$ $show that (i) angle projected 60^{°}$ $(ii) t_{1} + t_{2} = 2 \sqrt{\frac{3 h}{g}}$
Commented by Tawa11 last updated on 21/Sep/21

$nice sir$
	Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21

	$(i) For horizontal motion between both walls;$ $x = v_{x} t . But x = 2 h, v_{x} = 2 \sqrt{g h} \cos ϑ, t = 2 \sqrt{\frac{h}{g}}$ $\Rightarrow 2 h = (2 \sqrt{g h} \cos ϑ) (2 \sqrt{\frac{h}{g}})$ $\Rightarrow \frac{1}{2} = \cos ϑ \Rightarrow ϑ = 60 °$
	Commented by peter frank last updated on 20/Sep/21

	$thanks$
	Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21

	$(i i) For vertical motion :$ $y = u t + \frac{1}{2} {a t}^{2}, y = h, u = 2 \sqrt{g h} \sin 60 ° = \sqrt{3 g h}, a = - g$ $\Rightarrow h = \sqrt{3 g h} t - \frac{1}{2} {g t}^{2} \Rightarrow {g t}^{2} - 2 \sqrt{3 g h} t + h = 0$ $t = \frac{2 \sqrt{3 g h} \pm \sqrt{12 g h - 4 g h}}{2 g} = \frac{2 \sqrt{3 g h} \pm 2 \sqrt{2 g h}}{2 g}$ $t_{1} = \sqrt{\frac{3 h}{g}} + \sqrt{\frac{2 h}{g}}, t_{2} = \sqrt{\frac{3 h}{g}} - \sqrt{\frac{2 h}{g}}$ $t_{1} + t_{2} = 2 \sqrt{\frac{3 h}{g}}$
	Answered by peter frank last updated on 20/Sep/21

	Answered by peter frank last updated on 20/Sep/21

	$x - direction$ $x_{1} = v_{o} \cos θ t_{1}$ $x_{2} = v_{o} \cos θ t_{2}$ $v_{o} = 2 \sqrt{gh}$ $x_{2} - x_{1} = 2 h (t_{2} - t_{1}) = 2 \sqrt{\frac{h}{g}}$ $v_{o} \cos θ (t_{2} - t_{1}) = 2 h$ $\cos θ = \frac{2 h}{v_{o} (t_{2} - t_{1})} = \frac{1}{2}$ $θ = \cos^{- 1} \frac{1}{2} = 60^{°}$ $also y - direction$ $y = v_{o} \sin θ t_{1} - \frac{1}{2} {gt}_{1}^{2}$ $y = v_{o} \sin θ t_{2} - \frac{1}{2} {gt}_{2}^{2}$ $v_{o} \sin θ t_{1} - \frac{1}{2} {gt}_{1}^{2} = v_{o} \sin θ t_{2} - \frac{1}{2} {gt}_{2}^{2}$ $2 \sqrt{gh} \sin θ (t_{2} - t_{1}) = \frac{1}{2} g (t_{2}^{2} - t_{1}^{2})$ $2 \sqrt{gh} \sin θ = \frac{1}{2} g (t_{2} + t_{1}) [θ = 60]$ $t_{2} + t_{1} = \frac{4 \sqrt{gh}}{g} . \frac{\sqrt{3}}{2} =$ $t_{2} + t_{1} = 2 \sqrt{\frac{3 h}{g}}$
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