Solve this

some part of the motion may be related to projectile motion

Firstly i believe problem "overclaims"(i think problem needs some correction)!
(hypothesis of problem just means that coefficient of friction=tan@,@=angle of incline....i.e. net acceleration down the hill is 0)
Lets Take X-axis in "horizontal" direction and Y-axis down the hill!

X-axis situation
u_x(initial velocity)=1
a_x(acceleration)=-tan@g
so V_x=1- tan@g*t

Y-axis situation
u_y=0
a_y=0
so v_y=0

so speed would always be along horizontal and finally would become zero!

The book also has detailed solutions at the very end.

@vivek
thanks for the solution...but i stll am confused....Clearly even the solution say that there is no acceleration down the plane!further boy is pushed horizontally...i.e. to say his initial velocity down the plane is zero!so final velocity down the plane is zero of course.....where's the flaw?

If we replace mgsin(a) by k ,then force along x direction = \frac{dv_{x}}{dt} = -ksin\theta____(1) and force along y-direction is \frac{dv_{y}}{dt} = k-kcos\theta____(2)

From the figure in post 6,we see \frac{dy}{dx}= cot\theta\Leftrightarrow v_{y}= v_{x}cot\theta

differentiating \frac{dv_{y}}{dt}= \frac{dv_{x}}{dt}cot\theta- v_{x}cosec^2\theta\frac{d\theta}{dt}

plugging equation (1) and (2) in this relation ,we get k= - v_{x}cosec^2\theta\frac{d\theta}{dt}

combining this ,with equation (1) we get

\frac{dv_{x}}{v_{x}}= cosec\theta d\theta

on integrating with initial velocity(x-direction) 1m/s and final horizontal velocity V_x and initially \theta= \frac{\Pi }{2} and finally equal to \omega (Let) as limits of integration

we get,ln(V_{x}) = ln(tan\frac{\omega }{2})

or V_{x} = tan\frac{\omega }{2}

If the resultant velocity is V, then V = \frac{V_{x}}{sin\omega}\; =\; \frac{1}{(1+ cos\omega)}

we have got V as a function of the angle \omega
when \omega\rightarrow 0 ,V= \frac{1}{2}m/s