Question

a skier starts from rest and skis down a 82 meter tall hill labeled h1, into a valley and staught back up another 35 meter hill(labled h2). How fast in m/s is she going at the top of the 35 meter hill? Assume no friction

Answer 1

Answer:

She is going at 30.4 m/s at the top of the 35-meter hill.    

Explanation:

We can find the velocity of the skier by energy conservation:

$E_{1} = E_{2}$

On the top of the hill 1 (h₁), she has only potential energy since she starts from rest. Now, on the top of the hill 2 (h₂), she has potential energy and kinetic energy.

$mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2}$    (1)

Where:

m: is the mass of the skier

h₁: is the height 1 = 82 m

h₂: is the height 2 = 35 m

g: is the acceleration due to gravity = 9.81 m/s²  

v₂: is the speed of the skier at the top of h₂ =?

Now, by solving equation (1) for v₂ we have:

$v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m}$  

$v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s$    

Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.

I hope it helps you!