A particle is projected with velocity 2√(gh) so that it just clears two walls of height h which are at a distance of 2h apart from each other. Prove that the time of passing between the two walls is 2√(h/g).
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263Let the VELOCITY of the particle at the height h be u making angle θ with horizontal.clearly u=√(2gh).See that the maximum height of the projectile is 2h.so the particle will travel h distance upward and the corresponding range is 2h(given).treat this motion as a projectile. So (ucosθ)*T=2h; And T=2(usinθ)/g.put this value of T in the previous equation to get θ=45°.now calculate T.
Akash Anand @Sushovan: we had did this question as far as I remember.Upvote·0· Reply ·2014-06-04 02:01:54- ARKADYUTI BANDYOPADHYAY Then what? Equate the two equations in T?