Claculate the final speed of a skier who skis down a hill

lochs

I have an assignment question, it states: Surprisingly little advantage is gained by getting a running start in a downhill race. To demonstrate this, claculate the final speed of a skier who skis down a hill 80-m-high with negligible friction (a) if her initial speed is zero; (b) if her initial speed is 3.0 m/sec. [this final speed found in part (b) is larger than in part (a), but by far less than 3.0 m/sec!]

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Jameson

Use this kinematic equation:

$$v_{f}^2 = v_{i}^2 + 2ad$$

Compare the two.

xanthym

lochs said:
I have an assignment question, it states: Surprisingly little advantage is gained by getting a running start in a downhill race. To demonstrate this, claculate the final speed of a skier who skis down a hill 80-m-high with negligible friction (a) if her initial speed is zero; (b) if her initial speed is 3.0 m/sec. [this final speed found in part (b) is larger than in part (a), but by far less than 3.0 m/sec!]
For both cases:
{Delta Kinetic Energy} = -{Delta Potential Energy} = mgh = (9.81)(80)*m = (785)*m =
= m*(vf)2/2 - m*(vi)2/2 =
= (m/2)*{(vf)2 - (vi)2}
::: ⇒ (vf)2 - (vi)2 = (1570)
::: ⇒ vf = sqrt{1570 + (vi)2}

Case #1:
vi = 0 ::: ⇒ vf = sqrt{1570 + (0)2} = (39.6 m/sec)

Case #2:
vi = (3.0 m/sec) ::: ⇒ vf = sqrt{1570 + (3)2} = (39.7 m/sec)

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lochs

Thank you
xanthym, your response is a little confusing.. is that potential energy that you're using?

Jameson, i don't get how i'd get a? also would distance be 80 m? and if so isn't that height or are they the same thing?

Jameson

xanthym's right. I just glanced at this question... you don't have enough information to use a kinematic equation.

You know the potential energy at the top, and since their is no kinetic energy yet, you can call this total the total energy.

$$P.E. = mgh$$
$$= (m)(9.81)(80)$$

It is interesting to note that her mass does not affect this problem, as it will cancel out because it is used in every term.

Look at xanthym's work, and see how using the Law of Conservation of Energy will help you figure out the final velocities.

--------
Jameson

lochs

Ok so then
its PE=gh since mass cancels out?

this is what i have

KEi=PEi =KEf + PEf
becomes 0 + mgh inital = 1/2mv final squared + 0

v final=mgh inital = 2gh initial

v final= (2gh inital) 1/2

sorry i hope that makes sense.. i'm not sure how to use the latex thing..

Jameson

$$(2gh + v_i^2)^\frac{1}{2}$$

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lochs

okay i got it =)

thank you so much!

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