A particle is projected from a point A with velocity `usqrt2` at an angle of `45^(@)` with horizontal as shown in the figure. It strikes the plane BC at right angles. The velocity of the particle at the time of collision is

Options

A. `(sqrt3u)/2`
B. `u/2`
C. `(2u)/sqrt3`
D. `u`

Answer

C

Explanation

Let `v` be the velocity at time of collision.Then, `usqrt2 cos45^(@)=v sin 60^(@)`
`(usqrt2)(1/sqrt2)=sqrt(3v)/2 :. V=2/sqrt3u` (or)
At the moment of collision `tan 60^(@)=u/(("gt"-u))=sqrt3`
`v=sqrt(u^(2)+("gt"-u)^(2))=(2u)/sqrt(3)`