A particle is projected with velocity 2 gh so that it just clears two walls of equal height h , which are at a distance of 2h from each other. What is the time interval of passing between the two walls?

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Correct answer: D. 2 (h)/(g)

Let t be the time interval. Then, 2h = ( u _ x ) ( t ) or u _ x = (2h)/(t) ...(i) Further, h = u _ y t - (1)/(2)gt^ 2 or gt^ 2 - 2 u _ y t + 2h = 0 t _ 1 = 2 u _ y + 4 u _y^ 2 - 8gh 2g and t _ 2 = 2 u _ y - 4 u _y^ 2 - 8gh 2g t = t _ 1 - t _ 2 = 4 u _y^ 2 - 8gh g or u _y^ 2 = g^ 2 (t)2 4 + 2gh ...(ii) Given, u _x^ 2 + u _y^ 2 = (2 gh )2 4h^ 2 (t)2 + g^ 2 (t)2 4 + 2gh = 4gh (g2)/(4)(t)4 - 2gh(t)2 + 4h^ 2 = 0 (t)2 = 2gh ± 4g^ 2 h^ 2 - 4g^ 2 h^ 2 g^ 2 / 2 = (4h)/(g) or t = 2 (h)/(g)

A particle is projected with velocity 2 gh so that it just clears...

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