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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.

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"

Fc=mv^2/r

d=v1t+(1/2)at^2 (vertical)

v=d/t(horizontal)

Fc=mv^2/r

9.8mtan(60)=(mv^2)/2

9.8tan(60)=v^2/2

v=5.83m/s

This is where I think I am going wrong

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

t=.6389

v=5.83m/s

d=3.7 m?!

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

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..

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