A particle is projected from point A on the ground at an angle `theta` with horizontal. Another particle is projected from point B simultaneously from height `4 H` above point A with same velocity A and B are in the same vertical plane, where H is the maximum height attained by the particle A, Both the particle collide at the same time on the ground. Find the angle at which particle B is projected with hoizontal.

Options

A. `theta` upwards
B. `theta` downwards
C. zero
D. None of these

Answer

C

Explanation

`T_(A) = (2v sin theta)/(g) = (2 sqrt2gh)/(g) = sqrt((2g 4H)/(g))`
`( :. H = (u^(2) sin^(2) theta)/(2g))`
`:. T_(A) = T_(B) = sqrt(2 xx ((4H))/(g))`
`rArr (u_(B))_(y) = 0`
Hence particle B is projected horizontally.