A particle is projected along the line of greatest slope up a rough inclined plane at an angle of `45^(@)` with the horizontal. If the coefficient of friction is `(1)/(2)` then the retardation is :

Options

A. `(g)/(sqrt(2))`
B. `(g)/(2sqrt(2))`
C. `(g)/(sqrt(2))(1+(1)/(2))`
D. `(g)/(sqrt(2))(1-(1)/(2))`

Answer

c

Explanation

Here `a=g sin theta+ mug cos theta`
`= g sin 45^(@) + mug cos 45^(@)`
`=(g)/(sqrt2)+(g)/(2sqrt2)=(g)/(sqrt2)(1+(1))/(2)`
(c) is the choice.