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Solution:

Case 1. When the body is lauched up:

Let k be the cofficeint of friction, u the velocity of projection and  \( \ell \)  the distance traversed along the incline. Retarding force on the block  \( =mg\sin \alpha +kmg\cos \alpha \)  and hence the retardation  \( =g\sin \alpha +kg\cos \alpha \) .

Using the equation of  particle kinematics along the incline,  \( 0={{u}^{2}}-2(g\sin \alpha +kg\cos \alpha )\ell \)

Or,  \( \ell =\frac{{{u}^{2}}}{2(g\sin \alpha +kg\cos \alpha )}\,\,\,\,\,\,\,\,\,(1) \)  and  \( 0-u-(g\sin \alpha +kg\cos \alpha )t \)  or,  \( u=(g\sin \alpha +kg\cos \alpha )t\,\,\,\,(2) \)

Using (2) in (1)  \( \ell =\frac{1}{2}(g\sin \alpha +kg\cos \alpha ){{t}^{2}}\,\,\,\,\,\,\,(3) \)

Case (2). When the block comes downward, the net force on the body  \( =mg\sin \alpha -kmg\cos \alpha \)  and hence its acceleration  \( =g\sin \alpha -kg\cos \alpha \) .

Let, t be the time required then,  \( \ell =\frac{1}{2}(g\sin \alpha -kg\cos \alpha ){{{t}’}^{2}}\,\,\,\,\,\,\,(4) \)

From Eqs. (3) and (4)  \( \frac{{{t}^{2}}}{{{{{t}’}}^{2}}}=\frac{\sin \alpha -k\cos \alpha }{\sin \alpha +k\cos \alpha } \)

But  \( \frac{t}{{{t}’}}=\frac{1}{\eta } \) (according to the question),

Hence on solving we get  \( k=\frac{{{\eta }^{2}}-1}{{{\eta }^{2}}+1}\tan \alpha =0.16 \).