Grade 12 Physics, centripetal force and projectile motion

[Interest]

Hello Physics Forum, I am new member and this is my first Thread.
While onto the problem. I have been stuck on this particular for well over a week. I do not get the excat answer and to me my solution seems logically.

Homework Statement

An Australian bushman hunts kangaroos with a weapon that consists of a heavy rock tied to one end of a light vine of length 2m. He holds the other end above his head, at a point 2m above the ground and swings the rock in a horizontal circle. the cunning kangaroo has observed that the vine always breaks when the angle measure between the vine and the vertical reaches 60 degrees. At what minimum distance from the hunter can the kangaroo stand with no danger of a direct hit? Answer: 3m
radius=2m Fc=Tx=tan(60)*mg("According to my free body"), "Vine is 60 degrees from vertical"

Homework Equations

Centripetal Force
Fc=mv^2/r
Projectile Motion
d=v1t+(1/2)at^2 (vertical)
v=d/t(horizontal)

The Attempt at a Solution

Centripetal Force
Fc=mv^2/r
9.8mtan(60)=(mv^2)/2
9.8tan(60)=v^2/2
v=5.83m/s
Projectile Motion
This is where I think I am going wrong
Vertical
Since he is spinning the projectile horizontally, when thrown there will be no vertical velocity
v=0
Earlier in the question it said that he was spinning it 2m off the ground
d=2 (the rock must travel 2m before landing)
a=9.8m/s^2
d=v1t+(1/2)at^2
2=4.9t^2
t=.6389
Horizontal
t=.6389
v=5.83m/s
d=3.7 m?!

Please Help, I have been stuck on this for quite a while..

 

[Dr.D]

If he swings the rock in a horizontal plane as stated, how does the angle of the vine ever reach 60 degrees with the vertical? Something is wrong with this description.

 

[Interest]

If he swings the rock in a horizontal plane as stated, how does the angle of the vine ever reach 60 degrees with the vertical? Something is wrong with this description.

Same thing I thought too!, I asked my physics teacher and he said imagine a spinning ball connected to a pole. The ball is spinning horizontal, and it can make a angle with the vertical pole. Hope that clears misunderstandings.

 

[Dr.D]

Are you saying that the vine describes a cone?

If that is the case, you better draw a very careful free body diagram, including gravity, tension in the vine, and NO other forces on the stone (there is no means for any other real force to act on the stone). Then work through the kinematics to get the acceleration of the stone, so that you can correctly write the components of F = m a in each direction.

 

[Interest]

Yes, i know, but didn't I equate Fc to the right amount because isn't Fc=mgtan(60), since it is the horizontal component of the tension and the vertical component of the tension is mg, is it not? Can you please tell the best way to post my free body pic.

Edit: I think I did the centripetal force part correctly, what I believe I am having trouble with is the projectile motion component.

 

[Delphi51]

I see two difficulties. First, the radius is not 2 m. It is 2*sin(60).
Second, the initial height above the ground is not 2 m. It is 2 - 2*cos(60).
The rest looks good!

 

[Interest]

OK! i think just made tremendous progress with Delphi51's post (BIG THANKS), but before I make my solution final, can someone please clarify if the initial velocity for the vertical component when the rock is thrown as projectile truly zero, because instead of assuming it was zero, i made it equal to 5.4/tan(60) ("since 5.4 is your horizontal speed of the stone, using the Fc formula"), because i presumed that the projectile was released at a angle of 60 degrees to the vertical rather than straight forward because if I do I get the right answer.

 

[Delphi51]

"swings the rock in a horizontal circle" means there is no vertical component to the velocity. I get a = 16.99, v = 5.42, time of flight 0.4515, distance 2.45.