# Particle projected at an angle to the horizontal.

• Feb 6th 2013, 04:19 PM
Furyan
Particle projected at an angle to the horizontal.
Hello,

I'm sorry that this is a bit of a rambling question. I'm having difficulty understanding how to assign the signs in these problems. The rest of the math I can do and I'm finding it really frustrating that I'm not understanding the signs on the vectors. I can get to the solutions but I feel like I'm fudging it and I'd really appreciate it if someone would look over my work and clarify the sign thing for me. Thank you.

The first part of the question, which I can do is, 'A particle is projected from point O with a speed u at an angle of $\alpha$ above the horizontal and moves freely under gravity. When the particle has moved a horizontal distance x, it's height above O is y.

Show that:

$y = x\tan\alpha - \dfrac{gx^2}{2u^2\cos^2\alpha}$

I was able to do this and I looked it up and found it's an expression of the equation of trajectory. It's the second part of the question that I'm having difficulty with, which is:

'A girl throws a ball form point A at the top of a cliff. The point A is 8 m above a horizontal beach. The ball is projected with a speed of $7 ms^-1$ at an angle of elevation of 45 degrees. By modelling the ball as a particle moving freely under gravity, find the horizontal distance of the ball from A when the ball is 1 m above the beach.

I can solve this by substituting into the equation given in the first part if take the displacement y to be -7 m, I also had u as -7, but since this was squared, the sign makes no difference.

Solving:

$-7 = x\tan(45) - \dfrac{9.8x^2}{(2)(-7)^2\cos^2(45)}$, gives the correct answer.

I also tried solving the same problem without using the equation from the first part but instead using the vertical motion to find the time and then the horizontal motion to find the distance.

Using $s = ut + \dfrac{1}{2}at^2$, for the vertical motion.

Taking g, (a), as positive, since it's down, I get the correct time if I take u, $7\sin45$, as negative and the displacement, 7 m, as positive.

Solving:

$7 = -7\sin(45) + \dfrac{1}{2}(9.8)t^2$, gives the correct time.

One of the things that's confusing me is the displacement, in one equation it's negative in the other it's positive. The main thing is that it took me for ever to get the signs right to get the correct answer. I had the equations correct but without the right signs it was impossible to get the correct answer. My concern is that I'm not really understanding the signs on the vectors and I would really appreciate some help understanding how to assign the signs.

Thank you.
• Feb 6th 2013, 04:43 PM
HallsofIvy
Re: Particle projected at an angle to the horizontal.
The problem is that you have't specifically set up a "coordinate system". That is, you have not stated exactly where x= 0, y= 0 will be and which directions are positive.

"'A girl throws a ball form point A at the top of a cliff. The point A is 8 m above a horizontal beach. The ball is projected with a speed of at an angle of elevation of 45 degrees. By modelling the ball as a particle moving freely under gravity, find the horizontal distance of the ball from A when the ball is 1 m above the beach.

I can solve this by substituting into the equation given in the first part if take the displacement y to be -7 m"
Do you mean -8 m? There is no "7" distance mentioned here. If so then you are taking y= 0 to be at the "horizontal beach" and, though it strikes me as strange, y increasing downward. In this case "1 m above the beach" would be y= -1.

What the signs are is pretty much your choice. It depends upon your choice of a coordinate system.
• Feb 6th 2013, 05:47 PM
Furyan
Re: Particle projected at an angle to the horizontal.
Thank you for your reply. Yes I think my difficulty is exactly as you have stated, in not setting up a 'co-ordinate system' and defining where x = 0 and y = 0 and which directions are positive. I have tried very hard to do this and am finding it very confusing. I know that when solving these problems the directions of all the vectors involved must be the same and it is this that I'm having difficulty with. In this case if I take the horizontal through A as being y = 0, then when the ball is 1 m above the beach, which is 8m below A, that's a displacement of 7m downwards, so s, the displacement, is -7. Is that correct? Substituting this into the given equation gives me the correct solution so I assume it is. However I had to work backwards from the answer to figure this out and so I don't know if I've got the signs on other vectors incorrect instead. What's not helping me figure it out is that when I work out the problem the other way, in order to get the correct solution for the time I have to take the displacement as positive 7m, as I have shown my first post. That leaves me at a loss to figure out whether I've got the sign on the displacement wrong or whether the sign on one or more of the other vectors is incorrect. I have spent a very long time looking at this problem and the fact that the difficulty seems only to do with getting the correct signs on the vectors is driving me crazy(Headbang).

• Feb 7th 2013, 04:44 AM
Furyan
Re: Particle projected at an angle to the horizontal.
Hello HallsofIvy

I understand it now thank you.

In this equation, $y = x\tan\alpha - \dfrac{gx^2}{2u^2\cos^2\alpha}$, I know that y is the height above y = 0 and that in the second part of the question 1 m above the beach is 7 m below y = 0, so the vertical displcement is -7.

Also with this equation, $7 = -7\sin(45) + \dfrac{1}{2}(9.8)t^2$, I have taken down as positive so now the displacement is 7 m and the initial velocity is negative.

That all makes sense now, sorry for being so dense.

Thank you for you time.