Answer :
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:
[tex] E_{1} = E_{2} [/tex]
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.
[tex] mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2} [/tex] (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:
[tex] v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m} [/tex]
[tex] v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s [/tex]
Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.
I hope it helps you!