A particle is projected with velocity 2√(gh) ,so that it just clears two wall of equal height h which are at a distance of 2h from each other.show that the time of passing between the walls is 2√(h/g)

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Yatin Sharma · Answer

dude, what is the angle between the particle and the horizontal? the time would only depend on the vertical component, and to compute that we need the angle of projection

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A particle is projected with velocity 2√(gh) ,so that it just clears two wall of equal height h which are at a distance of 2h from each other.show that the time of passing between the walls is 2√(h/g)

A particle is projected with velocity 2√(gh) ,so that it just clears two wall of equal height h which are at a distance of 2h from each other.show that
the time of passing between the walls is 2√(h/g)

A particle is projected with velocity 2√(gh) ,so that it just clears two wall of equal height h which are at a distance of 2h from each other.show that

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A particle is projected with velocity 2√(gh) ,so that it just clears two wall of equal height h which are at a distance of 2h from each other.show that the time of passing between the walls is 2√(h/g)

A particle is projected with velocity 2√(gh) ,so that it just clears two wall of equal height h which are at a distance of 2h from each other.show thatthe time of passing between the walls is 2√(h/g)

A particle is projected with velocity 2√(gh) ,so that it just clears two wall of equal height h which are at a distance of 2h from each other.show that

1 Answers

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A particle is projected with velocity 2√(gh) ,so that it just clears two wall of equal height h which are at a distance of 2h from each other.show that
the time of passing between the walls is 2√(h/g)