Speed and Kinetic Friction

[defaultusername]

Homework Statement

Sam, whose mass is 75 kg, stands at the top of a 12-m-high, 100-m-long snow-covered slope. His skis have a coefficient of kinetic friction on snow of 0.07. If he uses his poles to get started, then glides down, what is his speed at the bottom?

Homework Equations

v = √(2g(h-μ_k√(L²-h²)))

The Attempt at a Solution

It seems as simple as plugging in the values into the equation, but my result isn't anywhere near what it should be.
From my understanding:
g = 9.8
h = 12m
L = 100m
μ_k = 0.07

v = √(2(9.8)(12-0.07√(100²-12²))) = 152.36
152.36 m/s? That seems a bit much to me. The back of my book says that it is 9.9 m/s, but how do I get there?
What did I mess up on?

 

[TSny]

v = √(2g(h-μ_k√(L²-h²)))

This is a very specialized formula! (Is this sort of formula supplied for you on exams?)

v = √(2(9.8)(12-0.07√(100²-12²))) = 152.36

When I calculate the left side, I don't get 152.36. Make sure you are doing the calculation correctly using your calculator.

 

[defaultusername]

Yeah we have to memorize it. I have no idea where it came from lol

When I punch it into my calculator I get the same answer. So I tried breaking it down algebraically as follows.
I still ended up with the same weird number.

Am I plugging the values into the wrong places?

 

[TSny]

See if you can spot the error in going from the 3rd to the 4th line. "Order of operations" is important here.

 

[defaultusername]

Rather than multiply, I take the 11.93^th root of 9856?

When I plug it into my calculator I get the correct answer, but I am concerned that when I do it by hand I am going to mess it up. :/

 

[TSny]

How would you evaluate $9 - 2 \cdot 3$

You have a subtraction and a multiplication. Which operation should you do first?

 

[defaultusername]

Oh! Duh!
Thank you!