Problem relating to skateboarding down a ramp

[marcus1255]

Homework Statement

here is the problem:

skater is to roll down a ramp with a 35 degree angle from horiziontal. the ramp then levels off to horizontal, with a 4 foot drop to level ground. the winner is the skater who can jump the furthest after leaving the ramp. your task is to determine the height of the starting gate at the top of the ramp. the typical skater pushes off from start at a speed of 3.0 m/s. for safety reasons, a skaters speed should never exceed 40 km/hour. Don't take friction or air resistance into account.

Homework Equations

The Attempt at a Solution

I converted the 40 km/hr to 11 m/s. I then drew out a rough sketch of what this whole thing would look like, but now i am at a loss of what to do next.

 

[Delphi51]

You have a pretty good start! That 11 m/s is the key number.
I think you can do it with energy, and fairly simply because no energy is lost. Looks like the skater has potential and kinetic energy at the top and just kinetic at the bottom. If you write that out in symbols and put in the known numbers, you will be able to solve for the height.

 

[marcus1255]

You have a pretty good start! That 11 m/s is the key number.
I think you can do it with energy, and fairly simply because no energy is lost. Looks like the skater has potential and kinetic energy at the top and just kinetic at the bottom. If you write that out in symbols and put in the known numbers, you will be able to solve for the height.

Can I just plug anything in for the mass? There is no mass given in the problem and I believe that I need one to calculate potential energy.

 

[marcus1255]

I solved it. Thanks for the advice!

I got a height of 5.6 meters. Does that sound about right?

 

[Delphi51]

Yes, sounds good. The question is just a little unclear - do we have the limit of 11 m/s at ground level or at the 4 foot level? Anyway, about 5.7 m above whatever it is. Did you cancel the m's like this:
½mv² + mgh = ½mV²
½v² + gh = ½V²