Kinetic Energy, Momentum, Projectile Motion ( A real stumper )

[xXmarkXx]

Sure. (Read the other posts in this thread!)

i have been..i'm really confused.

now i have...
(12sin(theta))/9.8=(14.2)/12cos(theta) does that look ok??

 

[Luke1294]

Doc...i'm not going to lie, I cannot find the arithmertic error for the life of me.

When I multiply through by g, i wind up with 140.238=12cos24sin. So I factor out a 12, and divide by that... which brings me to 11.68=2 sin x cos x.

Could you please give me a pointer as to where I'm going wrong? I don't just want the answer.

 

[Doc Al]

Doc...i'm not going to lie, I cannot find the arithmertic error for the life of me.

...giving me $$14.31=\frac{12cos\theta*24sin\theta}{g}$$.

Do it step by step:
(1) multiply by g
(2) divide by whatever you need to to get 2 on the right

 

[xXmarkXx]

i got 23.2=sin2(theta), but i used 14.2 instead of 14.3

do you agree doc?

 

[Luke1294]

Hm. So, I was looking at that the wrong way for sure. That can definatly be written as $$140.238= 288cos\theta sin\theta$$, can't it? So, I divide both sides by 144...giving me $$.973875=2sin\theta cos\theta$$, or $$.973875=sin(2\theta)$$. I then divide the left by 2, take the inverse sin, and arrive at 29 degrees?

Am i error free?! Finally?!

 

[xXmarkXx]

Hm. So, I was looking at that the wrong way for sure. That can definatly be written as $$140.238= 288cos\theta sin\theta$$, can't it? So, I divide both sides by 144...giving me $$.973875=2sin\theta cos\theta$$, or $$.973875=sin(2\theta)$$. I then divide the left by 2, take the inverse sin, and arrive at 29 degrees?

Am i error free?! Finally?!

luke, how did you get 14.3 for x? doesn't 14.3 = the momentum of the rock?

 

[Doc Al]

Hm. So, I was looking at that the wrong way for sure. That can definatly be written as $$140.238= 288cos\theta sin\theta$$, can't it? So, I divide both sides by 144...giving me $$.973875=2sin\theta cos\theta$$, or $$.973875=sin(2\theta)$$.

Good.

I then divide the left by 2, take the inverse sin, and arrive at 29 degrees?

Take the inverse sine first, then divide by 2 to get the angle theta.

 

[Luke1294]

Doc,

Thank you so much for your help. This problem even stumped our professor- apparently, he was instructing students to use some "likely" values to make the problem much simpler. Once again, thank you for helping me work through it.To the OP-
We're in this class together, pal. Jus' sayin.

 

[Luke1294]

As I'm looking back through this problem, I'm not sure i understand one little part- Why is it that the denomenator is 4.5 +65? I thought I had a handle on it earlier, but it's escaping me.

The distance between you and the rock was given to be 15.2m, that is:
$$p_{rock}-p_{you}=15.2\to{p}_{rock}=\frac{65.0}{4.5+65.0}*15.2m$$

$p_{rock}$ is the sought RANGE of the rock.
Did you follow this?

 

[xaer04]

you can find
$$p_{you}$$

by using conservation of momentum. set initial velocities to zero, and solve for
$$v_{you}$$

since momentum is constant after the throw (lack of friction, etc), that implies time is constant, so the M/S of the velocity becomes simply meters, and it can be used as a ratio of displacements, or
$$p$$

(arildno meant positions, btw) values. then, you know
$$p_{rock}-p_{you} = 15.2 m$$

and you can substitute in for
$$p_{you}$$

and the rest is just solving for p_rock.

i think it's really interesting that you guys spent more time solving this than me... :) i assume you're solving it for y_o = y_final (the starting height was the same as the landing height)? i still think there's not enough information though (i guess I'm just stubborn), but i am going to set up equations... heh, I'm just a computer science major anyway... physics isn't my strong subject (if that isnt' obvious already)

 

[Luke1294]

Thanks for the help. Like I said, you and I are in there with good old Mellendorf together...

If you read a ways up, you can see we actually found an angle, and then found the velocity of the man. Problem solved :)

 

[denverdoc]

hey we need something to do.

Actually I was surprised this one went on so long, as it was pretty straightforward by recognizing that the absolute velocity of separation was 3 percent larger than that of the rock alone. I agree tho, that the Yo issue was never resolved. I assumed it was thrown two handed as one might a bowling ball off the ground--after all it was close to 10#. Could have been shot put I suppose if the rock of that mass was small enuf to palm..

 

[xaer04]

denverdoc: heh... i just don't like holes in my problems, i suppose :)

luke: you wouldn't happen to be one of the guys that sits in the back left corner?

 

[Luke1294]

No sir, I happen to be upfront next to the deciptively cute blonde girl/ex girlfriend.

 

[xaer04]

ouch:(

so... would you happen to have been in cook's wednesday afternoon chem 130 lab last semester?

 

[Luke1294]

Yessir, that would also be me.

 

[denverdoc]

denverdoc: heh... i just don't like holes in my problems, i suppose :)

luke: you wouldn't happen to be one of the guys that sits in the back left corner?

Nor do I--so can someone tell me how this 140# guy threw from ice, and at rest, a 3.5 liter object some 50'. Thats 2/3'rds the size of a bowling ball, a distance of 2/3'rds the world shotput record!
J

 

[xaer04]

lol... got to love physics homework problems:)