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

[Molecular]

Homework Statement

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

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 $$\alpha$$

Homework Equations

I know the answer to the question is supposed to be:
$$\frac{2V^{2}_{o}cos^2\theta}{g*cos\alpha} * (tan\theta - tan\alpha)$$

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?

 

[tiny-tim]

… 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?

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.