66Its around 70.5°. Or to be precise sinθ<√8/9.
What is the maximum angle to the horizontal at which a stone can be thrown and always be moving away from the thrower ?
the interesting thing is all over projectiles i never thought of this as a problem ;)
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nimarjeet the answer is close to 70.11 or theta= sin inverse 2rt2/3.Upvote·0· Reply ·2019-07-12 22:51:58
29 Answers
1hehe 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
1Alternate 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 .
1Let 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
1see .......
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 !!
1ds2/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
106now i understood the question properly.... :( after such a long time
1oh 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 )
62kalyan you are right it does not :) ;)
but both are the same way to look at a similar problem :)
1@ Asish
even if it is lim(θ→π/2)θ
It is sure to come closer[3][3]
1I got the derivative as
ds/dt={u2cos2θ+(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]
1@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
106that'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
1Of course its not π/2 as , if it were π/2 the object comes back after attaining maax height
106i didnt get what the question is here "and always be moving away from the thrower"
62One suggestion which will help in the exams a lot!!!
if you have to check that ds/dt>0
then simply check
d(s^2)/dt greater than zero...
here because s is always positive!! :)
that will make calculations a bit easier... and t;he expression less clumsy..
11this is a question to which a variety of questions cn be asked
is air resistance there?[3]
1Solving this gives
θ>sin-1gt/u and
θ>sin-1(t2g/2u)
But I think I am stuck here
1The 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
11maybe 90° or just less than that
what i make of the question makes me tell this
1ok ok
i'll cut it short
THE PROBLEM STATEMENT IS C O M P L E T E