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 (sqrt(3)/(2))`
B. `45^@`
C. `60^@`
D. `30^@ + tan^-1(2 sqrt(3))`

Answer

A

Explanation

(a) `t_(A B)` = time of flight of projectile
=`(2 u sin (prop - 30^@))/(g cos 30^@)`
Now component of velocity along the plane becomes zero at point `B`.
`0 = u cos (prop - 30^@) - g sin 30^@ xx T`
or `u cos(prop - 30^@) = g sin 30^@ xx (2 u sin (prop - 30^@))/(g cos 30^@)`
or `tan(prop - 30^@) = (cot 30^@)/(2) = (sqrt(3))/(2)`
or `prop = 30^@ + tan^-1 ((sqrt(3))/(2))`,