# Tossing an item up a slope, calculating it's length from start to landing

1. Aug 21, 2008

### Molecular

1. The problem statement, all variables and given/known data
Right so I have the following scenario: An archer shoots an arrow upward a slope of angle $$\alpha$$, the angle between the arrows trajectory and the horizontal plane is denoted
$$\theta$$
The arrows speed in the direction in which it is fired is denoted $$V_0$$
as shown in this crudely drawn picture:

http://img124.imageshack.us/img124/8735/arrownb4.jpg [Broken]

My task is to find the formula for the length between the arrows initial position to the position where it lands, for all $$V_0$$, $$\theta$$ and $$\alpha$$
2. Relevant equations
I know the answer to the question is supposed to be:
$$\frac{2V^{2}_{o}cos^2\theta}{g*cos\alpha} * (tan\theta - tan\alpha)$$

3. The attempt at a solution
Well I started off by thinking that the arrows position measured by time for x and y coordinates respectively are:

$$x(t) = V_{o} * cos\theta * t$$

$$y(t) = V_{o} * sin\theta * t - \frac{1}{2} * g * t^2$$

I went on to show that as a function of x, y can be written:
$$y(x) = \frac{V_0y}{V_0x}x - \frac{g}{2v^{2}_{ox}}x^2$$
Since I know that on this items trajectory it has to stop where it intersects with the line $$y(x) = \alpha x$$

I put:
$$\alpha x = \frac{V_oy}{V_0x}x - \frac{g}{2v^{2}_{ox}}x^2$$

Now, replacing x with $$V_o cos\theta t$$ and $$V_oy$$ and $$V_ox$$ with their respective formulas of t and sin/cos $$\theta$$ I eventually wound up with the following expression of t

$$t = \frac{(tan\theta - \alpha)2cos\theta}{g}$$

and further more figured that since the length from the starting position to the position where the arrow lands is $$\sqrt{x^2 + y^2}$$ and I now have an expression for t that only uses constants, and I also know formulas for the position of x and for the position of y, I could replace the t in these formulas and put them in the root above to gain an expression for the length L. Such as this:

$$x(t) = V_{o} * cos\theta * t$$
$$y(t) = V_{o} * sin\theta * t - \frac{1}{2} * g * t^2$$
$$t = \frac{(tan\theta - \alpha)2cos\theta}{g}$$
$$L = \sqrt{x^2 + y^2}$$
$$L = (\sqrt{V_{o} * cos\theta * \frac{(tan\theta - \alpha)2cos\theta}{g})^2 + (V_{o} * sin\theta * \frac{(tan\theta - \alpha)2cos\theta}{g} - \frac{1}{2} * g * (\frac{(tan\theta - \alpha)2cos\theta}{g})^2 ) ^2}}$$

However as most of you can probably observe this is a right mess and most likely won't lead anywhere at all, and i'm willing to bet my left leg there's something quite easy I should be able to observe in order to solve the problme, I just can't seem to put my finger on it. Anyone willing to give me a nudge in the right direction would be greatly appreciated.

So anyone have any better idea on how to solve this?

Last edited by a moderator: May 3, 2017
2. Aug 21, 2008

### tiny-tim

Hi Molecular!

Yes, your proof looks fine.

(though you should be using tanα instead of α, and you could have left out the y(x) formula, and just put y = x tanα in the y(t) and x(t) formulas )

To get L, I'd recommend simplifying by defining µ (say) = 2cosθ (tanθ - tanα)/g …

the formula should then look quite neat.