A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall. Find the angle of projection of ball.

Options

A. `tan^(-1) ((3)/(2))`
B. `tan^(-1)((2)/(3))`
C. `tan^(-1)((1)/(2))`
D. `tan^(-1)((3)/(4))`

Answer

B

Explanation

Given, `R = (u^(2) sin 2 theta)/(g) = 24` .........(i)

`:' y = x tan theta - (gx^(2))/(2u^(2) cos^(2) theta)`
`3 = 6 tan theta - (36 g)/(2u^(2) cos^(2)theta)` .....(ii)
From Eq. (i), `(g)/(u^(2)) = (sin 2 theta)/(24) = (sin theta cos theta)/(12)`
Substituting in Eq. (ii) we have
`3 = 6 tan theta - (3)/(2) tan theta = (9)/(2) tan theta rArr theta = tan^(-1) ((2)/(3))`