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!

Vivek, i have the pdf with 89 problems only..leave about solution..i don't even have the answer

Clearly the friction co-efficient is tan(a).The force A(due to friction) is mgsin(a) and B is also mgsin(a).

The velocity along x and y direction is V_x and V_y.

There resultant makes angle theta with V_x. where theta is the angle as shown in figure.

(to be continued..)

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