A particle is projected with a certain velocity at an angle `prop` above the horizontal from the foot of an inclined plane of inclination `30^@`. If the particle strikes the plane normally, then `prop` is equal to.

Options

A. `30^(@)+tan^(-1)(1/(2sqrt3))`
B. `30^(@)+tan^(-1)((sqrt(3))/(2))`
C. `60^(@)`
D. `30^(@)+tan^(-1)(2sqrt3)`

Answer

B

Explanation

`t_(AB)=`time of flight of the projectile
`(2u sin (alpha-30^(@)))/(g cos 30^(2))`
Now component of velocity along the plane becomes zero at point `B`.
`:. 0=u cos (alpha-30^(@))-(g sin 30^(2))T`
or `u cos (alpha-30^(@))=(g sin 30^(2))(2u sin (alpha-30^(@)))/(g cos 30^(@))`
or `tan(alpha-30^(@))=1/(2tan 30^(@))=sqrt3/2`
`:. alpha=30^(@)+tan^(-1)(sqrt3/2)`