the velocity at the maximum height of a projectile is half of its initial velocity u.its range on the horizontal plane is?

Asked by meet | 17th Sep, 2012, 04:11: PM

Expert Answer:

Let the horizontal and verticle components of velocity be ux and uy respectively
thus, u = (ux2 + uy2)1/2
at the maximum height the only component of velocity will be the horizontal component which is unacccelerated and hence its value is ux
thus 1/2u = ux
1/2(ux2 + uy2)1/2 = ux
or uy = 31/2ux
or uy/ux = 31/2
let angle of throw be o
thus tan o = uy/ux
thus o = 60 degree
and ux = 1/2 u
thus range = R = (u2 sin 2o)/g = (u2 sin 120)/g

Answered by  | 18th Sep, 2012, 11:35: PM

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