Question

A skier traveling at 36.7 m/s encounters a 16.8 degree slope. If you could ignore friction, to the nearest meter, how far up the hill does he go? How to set this up?

Answer 1

Use conservation of energy to figure out how high (vertically) he will travel.\[\underbrace{mgh}_{\text{potential}}=\underbrace{\frac{1}{2}mv^2}_{\text{kinetic}} \Rightarrow h=\frac{v^2}{2g}\]Then, just use the geometry of the system to figure out how long the hypotenuse of a right triangle is that has an angle of \(16.8^\circ\) and an opposite side of length \(h\).

Answer 2

Hmm not following that at all

Answer 3

Have you guys done energy yet?

Answer 4

Rather, what is confusing to you?

Answer 5

What yakeyglee is saying is to consider the energy at the beginning and at the end of the situation. Initially, the skier only has kinetic energy. However, as he moves up the hill, his kinetic energy is slowly converted into gravitational potential energy. Eventually he'll come to a stop when all the kinetic energy, the energy of motion, is converted into potential energy. Therefore, the kinetic energy he had at the beginning, (1/2)mv^2 is equal to the potential energy he has at the end, mgh.

mgh = (1/2)mv^2 --> h = (v^2)/(2g)

The geometry of the system refers to the right triangle you can make out of the hill. We want to find out the distance the skier traveled up the hill, which in this case is the length of the hypotenuse of the right triangle, which I'll call x. By basic trigonometry, you know that sinθ = opposite/hypotenuse, which in this case is sin16.8 = h/x. You already know what h is from before, so you just need to solve for x. Let us know what you get :)

Answer 6

i got it thanks