So , the conclusion i come to is that ,the answer is no (provided the height is till which the helicopter is taken to is neglegible compared to radius , which even though not given in the question seems a reasonable assumption )

eh??how is angular momentum conservation an explanation to my logic???

my explanation is basic kinematics..

@ feynmann:

im not talkin abt any external force @ all!!

if the man jumps on a bus he comesback to the same position if bus moves st..

but wat abt in a merry go round?? it has circular motion...a man jumping WILL NOT land on the same spot cuz he is give a tgt velocity but the belt does not (in the material sense)move in the tgt direction!! so in circular motion this jumping will not land the man on same position

Remember H<<R

so it is basically a one dimensional plane kinematics , not a circular one !!

But if u want to be veryyyyyyyyyyyyyyyy precise , then it has to be given a very careful thought . U have to solve the differential eqn of motion .

And then we can come to the exact solution .

Do u wanna try ?

ofcourse im talkin abt jumping to a high enuf height...otherwise wats the pt of discussing this topic???

if it was some negligible height, then anchit cud have jus jumped and got his answer!!

n im only interested in the logic and reasoning not interested in solving..

and im not trying to be "veryyyyyyyyyyyyyyyy precise".. im jus talkin of the fundas..and i fail to understand y an expert is trying to challenge me to solve some complex diff eqn.. if v students cud do all tht then wat r u experts there for???

no offence to any expert..

@ computer .

Let's be serious !!

First of all , the analogy by which u are drawing conclusion is ENTIRELY FALSE . That's because in the merry - go-round u don't have any appreciable gravitational force towards it . But in the problem u HAVE TO CONSIDER the gravitational force on the helicopter or whatever .

So now explain ur reasoning .

I think  now it would be clear to u why I was telling to solve the differential eqn of motion . Got it ?

Now the problem can be formulated as follows:

Consider a ball given an upwards velocity v wrt earth which is rotating at a constant angular speed w. Find the time when the ball hits the Earth again .Also find the angular displacement of the ball on hitting the earth .

If earth moves by same amount by the same time then it will land at the same point , otherwise not .

Do u agree with this formulation ? If agree then plz go ahead using spherical polar co-ordinates . That's not  so difficult but may be time-consuming ,

Anyway , I am pleased with ur clear-cut comment . An academic comment , I believe( Unlike some experts ),should be made like this .

due to inertia, we also rotate along with the earth

so.... u wont reach america

Anyway ,after some calculation in polar co-ordinates I have obtained the final differential eqn in the formulated problem as follows:

The initial conditions are at t=0, r=R and

where R is the radius of the earth , M = mass of the earth and v is the velocity at which it was initially thrown up into the air ,wrt earth .

Now we have to solve this eqn to find the non-zero time when r=R .

And then proceed in the way in my previous post .

See, the differential eqn is not difficult itself but the calculations are a bit lengthy .

To solve it , u may put p = dr/dt

so LHS = pdp/dr

Now integrate both sides wrt r .

Then find p ( = dr/dt).

Then solve r(t) .( all the initial conditions are specified )

Please someone try it !!!