Distance Covered by Particle from A to +1 on Curve y=ax2

[The legend]

Homework Statement

A particle is released from a curve y=ax²(a=1m^-1) from A on smooth surface. It slides down and flies out from +1 on +tve x . find total horizontal distance D traveled by it .
(see link for image)
http://img59.imageshack.us/img59/6295/sumaj.png

Homework Equations

Equations of motion
LCE

The Attempt at a Solution

y1 = 1 * -2² = 4
y2 = 1 * 1 = 1
I used conservation of energy here.
mg * 4 = m*g*1 + 1/2 m * u²

Then got u = (60)^1/2

Again applying LCE i get
mg * 4 = 1/2 * m * v²

From this v = (80)^1/2

Now using v² = u² + 2*g*s

80 = 60 + 20s
s=1 metre

Adding 1 + {1 -(-2)}
= 4 = D

But that's not the actual answer. Any suggestions and can you please point out my mistake?

 

[kjohnson]

The first part to calculate the velocity at x=1m is correct. From that point you need to use your equations of motion and treat it like a projectile problem.

 

[The legend]

what about the components of velocity ... any hint on how to find them?

 

[kjohnson]

If this is a calc based class just take the derivative of ax^2 and evaluate it at x=1m, that will give you the slope at that point. Since slope is equal to y/x what trig function is equal to that as well?

 

[The legend]

oh right ... thanks!