a good one .

lets see after getting the fbd rest is just kinematics!!

but rohan on what basis are u saying its downward for any other angle ?????

my solution ultimately leads to your equation celestine but one step you missed as i shall show

see that just before stopping , V->0
Hence V1^2 +V2^2 ->0

thus
Vx->0 and Vy->0

hence v1/v2=d(v1)/d(v2) (By L hospital's rule) = (md(v1)/dt)/(md(v2)/dt)=F1/F2
where F1 and F2 are the forces on the object in the respective directions

but v1/v2=dy/dx=tana

hence writing the differential equations and solving for a
we get a as pi/2 or -(pi/2)

hence the body either moves in the upward direction or in the downward direction before stopping

uppward- when we give initial velocity along y only
downward - at any other angle ..

let v be at angle @ with downward vertical jus b4 stopping

let F be friction

mvcos@ = (Fcos@ -mgsinθ)dt

mvsin@=Fsin@dt

if sin@≠0

Fdt=mv

putting in first eq

we get mgsinθdt =0 implying dt=0 which is wrong !!

ok rohan im posting a very satisfying solution soon :)

rohan in this case net force is 0

i wasnt telling in general

and just before stopping friction may not balance mgsintheta upwards

it may act in a definite direction depending on velocity

before stopping let vx=a vy=b then a and b both are very less how can you say that frction direction is opposite to mgsintheta!!!!!

hey celestine ; how can you say that net force is zero when it stops !!!!

for example take a spring when it stops is the net force zero ???

v_y=gsinθ - μgcosθ
v_x=μgcosθ

v_x=0 when t₁=(v₀cosβ)/(μgcosθ)
v_y=0 when t₂=(v₀sinβ)/(gsinθ-μgcosθ)

if t₁>t₂ - velocity before it stops is down the plane

not able to come to a definite conclusion

ok by logic its got to be downwards

see after stopping net force is 0

jus b4 stopping friction balances mgsinθ down the plane approximately

ie Friction up plane this is possible when v down plane

oops totally forgot abt this Q
k im trying
u may expect a sol in 2-3 days :)

time to arise .... shall post solution soon if no answers .

ok i ll give a greater treatment to this after april26
( by logic its easy to say)

friction always opp velocity and also gravity acts down

so vertical vel is reduced more than hor vel so wen vertical vel is 0 , there is some hor vel now again hor vel is reduced but vertical vel inc initially and then somewhat opposed in later stages

continues till hor vel 0 and then only vertical vel down remains

its the same figure as the simple yet good one

instead of throwing horizontally it is thrown along the plane at an angle β to the horizontal

can you post a figure?

Posted just to make the headlines again !