How High Should a Ski Jump Ramp Be for Optimal Safety?

[clope023]

[SOLVED] Height of a Ski Ramp

Homework Statement

You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height h from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 2.0 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 30.0 m/s when they reach the bottom of the ramp. You determine that for a 80.0 kg skier with good form, friction and air resistance will do total work of magnitude 4000 J on him during his run down the slope.

Homework Equations

v1 = 2m/s
v2 = 30m/s
Ugrav1 = mgh
Ugrav2 = 0
K1 = 1/2mv1^2
K2 = 1/2mv2^2
Wother = 4000J

K1 + Ugrav1 + Wother = K2 + Ugrav2

The Attempt at a Solution

(1/2(80)(2^2)) + (80(9.8))h + 4000J = (1/2(80)(30^2))

h = ((1/2(80)(30^2))-(1/2(80)(2^2)) - 4000J)/(80(9.8)) = 41m wrong

not sure what I did wrong, I'm sure I'm using the right equations, any help is appreciated.

 

[qspeechc]

K1 + Ugrav1 = K2 + Ugrav2 + Wother
It should be as I have it.

Why is this so? Because Wother is lost on the way down, and hence cannot be part of the kinetic and potential energy at the bottom.

 

[clope023]

K1 + Ugrav1 = K2 + Ugrav2 + Wother
It should be as I have it.

Why is this so? Because Wother is lost on the way down, and hence cannot be part of the kinetic and potential energy at the bottom.

hmmm, didn't think of that, thanks.

so using that notation I received h = 51m, does that seem right?

 

[qspeechc]

I don't use calculator for this, you should try it, unless you already do.

80(2^2)/2 + 80(9.8)h = 80(30^2)/2 + 4000
2 + 9.8h = 450 + 50
h = 498/9.8 (now use calculator :D)
h = 50.82m
Yes, our answers agree. It seems reasonable to me, if you look at real olympic ski ramps, and take into account the energy lost etc.

 

[clope023]

I don't use calculator for this, you should try it, unless you already do.

80(2^2)/2 + 80(9.8)h = 80(30^2)/2 + 4000
2 + 9.8h = 450 + 50
h = 498/9.8 (now use calculator :D)
h = 50.82m
Yes, our answers agree. It seems reasonable to me, if you look at real olympic ski ramps, and take into account the energy lost etc.

thanks for the help, amazing how just the position of a variable can affect the whole problem, thanks again.

 

[qspeechc]

My pleasure. Yes, with energy problems it's particularly tricky, but alas, that is the only way to solve them >D. Thinking about what energy we start with (we didnt start with the 4000J) and what we end with, conservation and all that, usually helps.

 

[ej77]

I got a question from what book is this problem . Thanks