On an inclined plane two particles `A and B` are projected with same speed at the same angle with the horizontal, particle `A` down and particle `B` up the plane. If the ratio of time of flight of `A and B` is `cot theta`, where `theta` is the angle at which `B` is projected measured from inclined plane, find the angle at which particles are projected.

Options

A. `30^@`
B. `45^@`
C. `15^@`
D. `60^@`

Answer

B

Explanation

Time of flights of `A and B`
`T_A = (2 u sin (prop + beta))/(g cos beta)` and
`T_B = ( 2 u sin (prop - beta))/(g cos beta)`
`(T_A)/(T_B) = (sin(prop + beta))/(sin(prop - beta)) = cot theta` (given)
`rArr (cos theta)/(sin theta) = (sin (90 - theta))/(sin theta) = (sin (prop + beta))/(sin(prop - beta))`
We can write, `prop + beta = 90 - theta` ...(i)
and `prop - beta = theta` ...(ii)
on solving (i) and (ii) we get, `prop = 45^@`.
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