How far from the starting point do the pieces fall on ground

[annalian]

Homework Statement

An object is thrown with speed 20 m/Sunder the angle pi/3 rad with the horisontal direction. In the highest point, the object is divised into two pieces with same weights. One of them, after the division has the speed 0.
How far from the starting point do the pieces fall on ground?

Homework Equations

t=2v0sina/g

The Attempt at a Solution

The piece that has the speed 0:
x1=v0^2sin2a/2g=17.32 m
x2=v0^2sin2a/g=34.64 m
The answer in the book is 10, 20 meters.

 

[PeroK]

Homework Statement

An object is thrown with speed 20 m/Sunder the angle pi/3 rad with the horisontal direction. In the highest point, the object is divised into two pieces with same weights. One of them, after the division has the speed 0.
How far from the starting point do the pieces fall on ground?

Homework Equations

t=2v0sina/g

The Attempt at a Solution

The piece that has the speed 0:
x1=v0^2sin2a/2g=17.32 m
x2=v0^2sin2a/g=34.64 m
The answer in the book is 10, 20 meters.

This problem is in two parts. Can you describe what happens in each part?

Because of this, you will need to be careful about what you mean by $v_0$ and $t$. You will have to define all your variables carefully before you use any equations to relate them.

PS In fact, it's in three parts!

 

[annalian]

This problem is in two parts. Can you describe what happens in each part?

Because of this, you will need to be careful about what you mean by $v_0$ and $t$. You will have to define all your variables carefully before you use any equations to relate them.

The first piece follows the movement as if nothing happened (as if it was the object without being devised.)
That's why for it I used L=v0^2sin2a/g
v0=20 m/S is the initial speed of the object
a-the angle the object forms with the horizon.
The second piece just does half of the complete movement, as when it reaches the top, the speed becomes 0. So the road is half of the one of the first piece.

 

[PeroK]

The first piece follows the movement as if nothing happened (as if it was the object without being devised.)
That's why for it I used L=v0^2sin2a/g
v0=20 m/S is the initial speed of the object
a-the angle the object forms with the horizon.
The second piece just does half of the complete movement, as when it reaches the top, the speed becomes 0. So the road is half of the one of the first piece.

I think your answer is better than the one in the book. Do you think conservation of momentum might be involved?

The book answer can't possibly be correct for the problem as stated.

 

[annalian]

I think your answer is better than the one in the book. Do you think conservation of momentum might be involved?

The book answer can't possibly be correct for the problem as stated.

No, we haven't repeated momentum yet. Do you think the answer will be right, by using the momentum?

 

[PeroK]

No, we haven't repeated momentum yet. Do you think the answer will be right, by using the momentum?

I would perhaps leave this question. The book answer is clearly not right. But, you might ask yourself why it said the two pieces were of equal weight? Why is the size of the two pieces important?

 

[annalian]

I would perhaps leave this question. The book answer is clearly not right. But, you might ask yourself why it said the two pieces were of equal weight? Why is the size of the two pieces important?

to use the expression mv=m1v1+m2v2
v=v1+v2, v1=0
v2=v

 

[PeroK]

to use the expression mv=m1v1+m2v2
v=v1+v2, v1=0
v2=v

Shouldn't that be $m_1 = m_2 = \frac{m}{2}, \ \ v_2 = 2v$?

 

[annalian]

Shouldn't that be $m_1 = m_2 = \frac{m}{2}, \ \ v_2 = 2v$?

Yes, you're right

 

[annalian]

Yes, you're right

So in this case, the road for the first one would be the same, but for the second, I would have to add to the first one v0^2sin2a/2g, where v0=40 m/s?

 

[PeroK]

So in this case, the road for the first one would be the same, but for the second, I would have to add to the first one v0^2sin2a/2g, where v0=40 m/s?

No, you need to think more carefully about the horizontal speed of the second piece. It is $v$ until the highest point, and $2v$ after that.

 

[annalian]

No, you need to think more carefully about the horizontal speed of the second piece. It is $v$ until the highest point, and $2v$ after that.

Yes, I said it is 20^2sin2a/2g+40^2sin2a/g. Is this correct?

 

[PeroK]

Yes, I said it is 20^2sin2a/2g+40^2sin2a/g. Is this correct?

Yes, of course, that's correct.

I was thinking that simply it goes 3 times as far as the first piece! I wasn't thinking in terms of $g$ and angles and things.