Physics - Projectile motion

With reference to the diagram below:

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2007-09-27 09:30:14 補充：
Yup, pls wait for a moment.

2007-09-27 10:11:47 補充：
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Hei man
The way suggested above is a good way

If ur mathematics is good , u can also try the following way.
In this case , i will only consider the original x and y direction

we first write down the coordinate of the ball as a function of time
let A be alpha, P be phi, x be x-coordinate , y be-y coordinate
x = vo * t * cos(A) --------------------------(1)
y = vo* t *sin(A) -1/2(g)t^2----------------(2)

once the ball hit the sloped ground
the x and y coordinates of the ball is actually on the line of the ground
the equation of this line is :
Y=MX+C
Y=MX (since we can set the line pass through origin,hence the y-intercept is 0)
y=tan(P)x---------------------------------------(3) (slope is tan(P))

now we have
x = vo * t * cos(A) --------------------------(1)
y = vo* t *sin(A) -1/2(g)t^2----------------(2)
y=tan(P)x---------------------------------------(3)

by solve (1),(2),(3) we can get the x coordinate when the ball hit the ground
and idea to solve it is to eliminate y and t since other letter are known
arrange (1) to t=x/(v0cosA) sub this together with (3) into (2)
we will have a relation with x and other constant only
hence it is easy to solve x

after some elimination
X=2[tan(A)-tan(P)](v0cosA)^2/g

2007-09-27 19:47:27 補充：
X is only the x-direction displacement it is not the range range is the displacement including x-y directionthe range or displacement x/R=cos(P)hence Range = x /cos(P)=2[tan(A)-tan(P)](v0cosA)^2/gcos(P)This is actually the same as the above answer but using different method haha

2007-09-27 19:50:04 補充：
May be the answer in ur hand have some problem