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:
<br /> R = \frac{(2{V_0}^2 cos \beta sin(\beta - \alpha )}{g {cos}^2\alpha}<br />

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 \alphaand \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
<br /> \triangle y = (V_0 sin \beta)\frac{cos \alpha}{V_0 cos \beta} - \frac{1}{2} g ( \frac{cos \alpha}{V_0 cos \beta} )

<br /> \triangle y= \frac {V_0 sin \beta cos \alpha}{V_0 cos \beta} - \frac{1}{2} g ( \frac{cos \alpha}{V_0 cos \beta})
<br /> sin \alpha = tan \beta cos \alpha - \frac {g {cos}^2 \alpha}{2 {V_0}^2 {cos}^2 \beta}<br />
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 \alphaand \triangle y = sin \alpha
...

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

\triangle x = Rcos \alphaand \triangle y = Rsin \alpha

 

[zanazzi78]

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

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

<br /> R Sin \alpha = R tan \beta cos \alpha - \frac {R g {cos}^2 \alpha}{2 {V_0}^2 {cos}^2 \beta}<br />

but i don`t see how this helps?

 

[Fermat]

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