Speed, distance and spring on a track

[brunettegurl]

Homework Statement

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 30^o on the other side. How far along the slope will Tarzan get?
Picture can be found here https://www.physicsforums.com/showthread.php?t=315325"

Homework Equations

e=0.5mv²
eg=mgh
spring = 0.5kx²

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 don't need to calculate initial velocity. Just use the spring potential energy.

 

[brunettegurl]

so then wld it look like mgh_initial+0.5*kx²= mgdsin$$\vartheta$$...where h_initial= 11m ??

 

[dx]

correct.

 

[brunettegurl]

thanks :)