speed, distance and spring on a track

[brunettegurl]

1. The problem statement, all variables and given/known data

Tarzan (m=90kg) presses against a horizontal spring with spring constant k=81000N/m ( on a frictionless surface), compressing it by 100 cm. After loosing contact with the spring, Tarzan starts going down an 11.0 m high hill that is frictionless, and then starts to go up a frictionless incline of 30o on the other side. How far along the slope will Tarzan get?
Picture can be found here http://www.physicsforums.com/showthread.php?t=315325

2. Relevant equations

e=0.5mv2
eg=mgh
spring = 0.5kx2

3. The attempt at a solution

this question very similar to a previous question i posted but i can not seem to carry it forward and find the distance the person would travel it they went up the hill of the frictionless hill. i first found the speed which was 30 m/s. since the track is frictionless i assummed the speed would be the same and set up and another equation 0.5mv2=mgh where h= dsin\vartheta and i solved for d this didnt work out ..pls. help

[Noein]

The speed wouldn't be the same all along the track. Tarzan accelerates down the hill and decelerates up the ramp.

[brunettegurl]

so do i find the acceleration of the person using a kinematic equation where the acclerration is gravity - x. x being the deceleration>>??

[dx]

Use conservation of energy. Acceleration is not needed here.

[brunettegurl]

if i used conservation of energy then wouldn't this be right 0.5mv2=mgh where h= dsin\vartheta

[dx]

You forgot the initial gravitational potential energy. Also, you dont need to calculate initial velocity. Just use the spring potential energy.

[brunettegurl]

so then wld it look like mghinitial+0.5*kx2= mgdsin\vartheta...where hinitial= 11m ??

[dx]

correct.

[brunettegurl]

thanks :)