How is the Range of a Projectile on an Inclined Plane Derived?

[zanazzi78]

Q. A projectile is fired with an initial speed $$V_0$$ at an angle $$\beta$$ to the horitontal. Show that it's range alnog a plane which it's self is inclined at an angle $$\alpha$$ to the horitontal $$( \beta > \alpha)$$ is given by:
$$R = \frac{(2{V_0}^2 cos \beta sin(\beta - \alpha )}{g {cos}^2\alpha}$$

A.So I`ve started off with
$$\triangle x = (V_0 cos \beta) t$$
and
$$\triangle y = (V_0 sin \beta) t - frac{1}{2} g t^2$$

$$\triangle x = cos \alpha$$and $$\triangle y = sin \alpha$$
so i rearranged $$\triangle x$$ to get $$t = \frac{cos \alpha}{V_0 cos \beta}$$ and sub it into $$\triangle y$$
now i have
$$\triangle y = (V_0 sin \beta)\frac{cos \alpha}{V_0 cos \beta} - \frac{1}{2} g ( \frac{cos \alpha}{V_0 cos \beta} )$$

$$\triangle y= \frac {V_0 sin \beta cos \alpha}{V_0 cos \beta} - \frac{1}{2} g ( \frac{cos \alpha}{V_0 cos \beta})$$
$$sin \alpha = tan \beta cos \alpha - \frac {g {cos}^2 \alpha}{2 {V_0}^2 {cos}^2 \beta}$$
now i`m stuck ... Any hint`s/tips would be great.
Cheers.

 

[Astronuc]

Somewhere there seems to be an 'R' missing.

Also, one of the latex expression needs \ in front of frac.

 

[daniel_i_l]

delta x = cos(alpha) ??

 

[zanazzi78]

delta x = cos(alpha) ??

alpha is the angle of the inclinded plane. it forms a right triangle with the horizontal and the point at which the projectile meets the inclinded plane. therefore delta x is equal to cos alpha.

 

[Fermat]

...
$$\triangle x = cos \alpha$$and $$\triangle y = sin \alpha$$
...

This is where the R is missing. They should be,

$$\triangle x = Rcos \alpha$$and $$\triangle y = Rsin \alpha$$

 

[zanazzi78]

This is where the R is missing. They should be,
$$\triangle x = Rcos \alpha$$and $$\triangle y = Rsin \alpha$$

ok so i`ve put the 'R' s in and get

$$R Sin \alpha = R tan \beta cos \alpha - \frac {R g {cos}^2 \alpha}{2 {V_0}^2 {cos}^2 \beta}$$

but i don`t see how this helps?

 

[Fermat]

You're almost there. But the R in the last term should be R²