Two inclined planes OA and OB intersect in a horizontal plane having their inclinations `alpha and beta` with the horizontal as shown in figure. A particle is projected from P with velocity u along a direction perpendicular to plane OA. The particle strikes plane OB perpendicularly at Q.

If `alpha = 30^@ , beta = 30^@` the time of flight from P to Q is

Options

A. `tan^(-1) [(P^(2)+PQ +Q^(2))/(PQ)]`
B. `tan^(-1)[(P^(2)+Q^(2)-PQ)/(PQ)]`
C. `tan^(-1)[(P^(2)+Q^(2))/(2PQ)]`
D. `sin^(-1)[(P^(2)+Q^(2)+PQ)/(2PQ)]`

Answer

A

Explanation

The equation of trajectory.
`y = x tan alpha [1-(x)/(R)]` gives
`Q = P tan theta [1-(P)/(R)]` ...(i)
and `P = Q tan theta [1-(Q)/(R)]` ......(ii)
On dividing we get, `(Q^(2))/(P^(2)) = ([1-P//R])/([1-Q//R])`
`(1)/(R) [P^(3)-Q^(3)] = P^(2)-Q^(2)`
`R = (P^(3)-Q^(3))/(P^(2)-Q^(2)) = (P+PQ +Q^(2))/(P+Q)`
Now, `(Q)/(P) = tan theta [1-(P(P+Q))/(P^(2)+PQ+Q^(2))]`
`= tan theta [(P^(2)+PQ+Q^(2)-P^(2)-PQ)/(P^(2)+PQ+Q^(2))]`
`rArr tan theta = (P^(2)+Q^(2)+PQ)/(PQ)`
`rArr theta = tan^(-1) [(P^(2)+PQ+Q^(2))/(PQ)]`