Swinging Rock Hunting: Allister's Deer Catch Method

[MikeH]

Swinging Vine(HELP ASAP)

All help is apprieciated.

Backwoods Allister hunts deer with the following weapon...a rock attached to a 2m vine. He holds the end above his head at a point 2m above the ground, when the above ground angle reaches 60 degrees with the vertical the rock breaks off and flies towards the deer. At what minimum distance from Allister can the deer stand with no danger of a direct hit?

 

[Jupiter]

I think it depends on how tall the deer is. But all in all this is a simple trig problem. Why don't you give it a shot first?

 

[MikeH]

Well here's what I have so far. I used trig to find that the rock will be 1m above the ground and that the radius of the circle is 1.7m. From here I'd need to find the speed? Then use projectile? I'm kinda lost here.

 

[NateTG]

Try applying conservation of energy.

 

[MikeH]

How so?

 

[NateTG]

The change in kinetic energy=the change in potential energy. That will give you the velocity of the rock. From there it's straight kinematics.

 

[MikeH]

I know KE = 1/2mv^2
I know that the change in GPE = mg(delta)h
How would one write the change in KE?
1/2mv^2 - 1/2mvf^2 ??

 

[MikeH]

Anyone?

 

[HallsofIvy]

Your original problem does not give enough information to answer this. You would need to know how fast he is swing the rock.

 

[MikeH]

There must be some way, my teacher wouldn't have given a question that wasn't possible to answer.

I need to work to this formula: dh = vht

To find t(time).
dv= 1/2gt^2
t = (square root of)(1/2gdv)
t = 2.21 seconds

Now if I try to use Conservation of Energy.

KE(i) + GPE(i) = KE(f) (There is no GPE since it will be on the reference point)
.5mv(i)^2 + mgh = .5mv(f)^2
Masses cancel
.5v(i)^2 + gh = .5v(f)^2

This give you any ideas?

 

[NateTG]

You'd make my life a whole lot easier if you used LaTeX, but that's not really something you can deal with right now.

Ok, so:
$$KE_0+PE_0=KE+PE$$

Now, the change in height is:
$$2mg\sin{60}=\sqrt{3}mg$$

So, we have:
$$\frac{1}{2}mv_{launch}^2=\frac{1}{2}mv_0^2+\sqrt{3}mg$$
Some cancelations give:
$$v_{launch}=\sqrt{v_0^2+2\sqrt{3}}g$$

Whatever the initial velocity that you're dropping the weight with is is $$v_0$$. You might see some hint for that somewhere.

From there you can apply the normal kinematic formulae.

 

[MikeH]

I don't understand what you mean. V(launch) is the speed that it leaves the rope and V(o) is the speed that it hits the ground with?

 

[NateTG]

$$v_0$$ is the speed at the top of the swing.

 

[MikeH]

I've been looking at this as swinging around him horizontally, I guess I've been looking at it the wrong way all along.

 

[MikeH]

How did you find the change in height?
$$2mg\sin{60}=\sqrt{3}mg$$


I'm not really following this question too well. The more I think about it the more I get confused. Hopefully I get a snowday tomorrow so I don't have to pass it in. I need the help ASAP!
Thanks

 

[NateTG]

Maybe that should be cosine. You should be able to use trig to find the change in height.