Conservation of energy / work problem

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adams_695
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Homework Statement
Call the system the bicycle and the rider. Use the work-energy equation and W = Fd. Assume the cyclist and air do not heat up. The work-energy equation is Kf + Ugf = Ki + Ugi + W.
Relevant Equations
Kinetic energy & gravitational energy
Handwritten worked solution on lined paper: Answer: vF^2 = vi^2 + 2gy - 2Fnc d/m, = (12 m/s)^2 + 2(9.8 m/s^2)(30 m) - 2(120 N)(450 m)/70 kg, vF = 24 m/s.
Handwritten work-energy derivation under the heading Attempt, solving for final speed v_F step by step from the energy equation with drag force, ending with v_F = square root of (v_i^2 - 2gy_f + 2gy_i - 2F_drag d / m).
Handwritten algebra attempt solving for v_f from the energy equation (1/2)mv_f^2 + mgy_f = (1/2)mv_i^2 + mgy_i - F_drag d, ending with v_f = sqrt(v_i^2 - 2g(y_f - y_i) - 2F_drag d / m).

If someone could advise what I've done wrong it would be much appreciated. How have they eliminated the initial and final for y, and simplify only to y? Also, how did they simplify to a positive 2? What algebraic steps have I missed? Thanks for your help.
 
I would not call that a problem statement. What is the actual question?
 
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haruspex said:
I would not call that a problem statement. What is the actual question?
Printed physics homework text showing the end of problem 7 about a plane landing on a spring, then problems 8 and 9: a frisbee stuck 16 m up in a tree, and a cyclist coasting down a 450-m slope.


Sorry here it is. Question 9. Thanks!
 
haruspex said:
You are told the top of the hill is 30m higher than the bottom of the hill. So what is yi-yf?

30m-0m.

Is this how I went wrong when solving algebraically in the image of my working?
 
There was nothing wrong in what you worked out. As haruspex pointed out, identify yi and yf in your own algebra and you are done.
 
adams_695 said:
30m-0m.

Is this how I went wrong when solving algebraically in the image of my working?
The last line of your working has the terms -2gyf+2gyi. Factorise.