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A particle is projected vertically upwardswith a velocity of 20 m/sec. Find thetime at which thedistance travelled is twice thedisplacement. |
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Answer» The maximum height H = v^2/ 2g = 20*20/(2*10) = 20m.Time to reach maximum height is t = v/g = 20/10 = 2 s.If distance = H + x, displacement = H-x.Given H+ x = 2*(H-x)=> x = H/3 = 20/3 m.Time to fall a distance of 20/3 m from rest is found fromt^2 = 2h/ g = 2*(20/3) / 10 = 4/3t = 2/ root3Total time = 2+ 2/ root3 s. |
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