very easy but interesting

Its around 70.5°. Or to be precise sinθ<√8/9.

hehe bhaiyya i think that should have struck naturally
if kalyan got as far as he did ,
he should have no problem in nailing this one

Alternate and more easily " grabbable " solution -

At any moment of time , the co - ordinates of the object undergoing projectile motion are -

x = x

y = x tan θ - g x ²2 V ² cos ² θ

Now , the distance from origin ( 0 , 0 ) to this point -

S ² = x ² + ( x tan θ - g x ²2 V ² cos ² θ ) ²

For " S " to be increasing , " S ² " must also be increasing since , " S > 0 " for all " x " .

In other words , dSdx > 0 ...............implies

dS²dx > 0

From here - on , after a bit of calculations , we easily arrive at the required result .

Let us assume that the path of the projectile starts from origin and bends like a down - ward parabola quite obvious . .

Let the position vector of the point which is furthest from origin be " r " and the velocity vector of the projectile at this position be " v " .

Obviously , this point is on that normal of the parabola which passes through origin .

Hence , ( r ) . ( v ) = 0

Since , r = x i + y j = V t cos θ i + ( V t sin θ - g t ²2 ) j

And , v = V cos θ i + ( V sin θ - g t ) j

Hence ,

t ² - 3 V sin θg t + 2 V ²g = 0

Since we do not want any roots of this equation , so , " Δ < 0 " .

From there , it is easily seen that -

sin θ < 2 √ 23

see .......
i dont quite get where are u messing it up....

the u term would get cancelled so will the t term
as both ≠0
but hey wait !! did u think in terms of quadratics surely u cannot solve without that
that Δ < 0 thing !!

ds²/dt = 2u²t+g²t³-3ugsinθt²

sinθ<=(2u²-g²t³)/3ug

minimum t=0

Hence (sinθ)_max=2u/3g

So it depends on u[7][7]

Another point of doubt

ThanQ philip

Got U
Doin it now[90]

now i understood the question properly.... :( after such a long time

oh kalyan instead of finding
ds/dt

find d(s²)/dt = 2sds/dt>0

since s > 0
2sds/dt>0 would imply ds/dt >0

that was wat we said ...
but i didn''t elaborate coz i thot it was "commonsense " for u hehe[1]

and now see if u get wat sir has given above
(u should now )

I got the derivative as

ds/dt={u²cos²Î¸+(usinθt-gt²/2)(usinθ-gt)}/√~

The √~ here is +ve, so it need not be considered 4 the inequation

So how does squaring help here[7][7][7]

@kalyan hehe [4] now i got it why u got stuck
kalyan if u'd asked me to differentiate that stuff in that form i'd plain refuse even if u had a pistol in hand (forgive exaggeration :D )

do as bhaiyya said ... and u'll be through in no time
.
.
.

as i did it this way i expected everyone to do this way
but..
there is one more method that gives a hell lot more insight into the stuff
i'll say if this one is mathematical that one is more elegant

that's y i said lim... not exactly π/2 .. even if the angle is just less than π/2 the ball will still move away from the thrower

Of course its not π/2 as , if it were π/2 the object comes back after attaining maax height

i didnt get what the question is here "and always be moving away from the thrower"

kalyan has almost done it
check for careless mistakes

does that mean he is right [34]

maybe 90° or just less than that

what i make of the question makes me tell this

ok ok
i'll cut it short
THE PROBLEM STATEMENT IS C O M P L E T E