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A particle is projected with velocity 2(...

A particle is projected with velocity `2(sqrt(gh))`, so that it just clears two walls of equal height h which are at a distance of 2h form each other. Show that the time of passing between the walls is `2 (sqrt (h/g)).` [Hint : First find velocity at height h. Treat it as initial velocity and 2h as the range. ]

Text Solution

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`s=ut+1/2 at^(2) rArr a=(u sin theta) t-1/2 g t^(2)`

`rArr t=(u sin theta +-sqrt(u^(2) sin^(2)theta-2ag))/g`
`Deltat=(2sqrt(u^(2) sin^(2) theta-2ag))/g`
For horizontal motion : `2a=u cos theta xx Deltat`
`rArr 2a=(u cos thetaxx2sqrt(u^(2) sin^(2) theta-2ag))/grArr theta=60^(@)`
`:. Deltat=(2a)/(u cos theta)=(2a)/(2sqrt(ag)xx1/2)=2sqrt(a/g)`
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