very easy but interesting

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

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 has almost done it
check for careless mistakes

this is a question to which a variety of questions cn be asked

is air resistance there?[3]

Solving this gives

θ>sin^-1gt/u and
θ>sin^-1(t²g/2u)

But I think I am stuck here

@subash[3]

lim(θ→π/2)θ ??

The displacement along x axis is x=ucosθ t
along y axis is y=usinθ t-gt²/2

distance from thrower, s= √(x²+y²)

This increases when ds/dt >0

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

you include obtuse angles as well?