A particle just clears a wall of height b at distance a and strikes the ground at a distance c from the point of projection. The angle of projection is (1) `tan^(-1)b/(a c)` (2) `"45"^o` (3) `tan^(-1)(b c)/(a(c-a)` (4) `tan^(-1)(b c)/a`

Answer

A

Explanation

`T_(1) sin theta = mg + T_(2) sin theta`
`200 (4//5) - mg = T_(2) sin theta`
`160 - 40 = T_(2) (4//5)`
`T_(1) cos theta T_(2) cos theta = mv^(2) r`
`(3//5) (350) = 4 xx (omega^(2)) (3)`
`omega^(2) = (70)/(4), omega = sqrt((35)/(2))` rad/sec
Rev./ min `= sqrt((35)/(2)) xx 60 xx (1)/(2 pi)` `~~ 39.6` rpm