A skier slides down a hill starting with zero velocity at a height of h1 = 45.2 m above the bottom of the hill. The shape of the terrain is like a cosine wave.
a) What is the velocity of the skier on top of the second, intermediate hill, whose height is h2 = 36.5 m? Neglect all frictional effects.
b) What is the skier's velocity at the bottom of the hill?
(1/2) m v^2 = m g h
or
v = sqrt(2 g h)
where h is how far down you moved
g is about 9.81 m/s^2
To find the velocity of the skier at different points on the hill, we can use the principle of conservation of energy. The total mechanical energy of the skier is conserved as they slide down the hill, neglecting friction. The total mechanical energy is the sum of potential energy and kinetic energy.
a) To find the velocity of the skier at the top of the second hill, we can use the conservation of energy principle.
The initial potential energy (P.E.) of the skier at the top of the first hill is equal to the final sum of potential and kinetic energy at the top of the second hill.
At the top of the first hill:
Initial potential energy: P.E.1 = m * g * h1 ...(1)
where m is the mass of the skier, g is the acceleration due to gravity, and h1 is the height of the first hill.
At the top of the second hill:
Final potential energy: P.E.2 = m * g * h2 ...(2)
Final kinetic energy: K.E.2 = (1/2) * m * v2^2 ...(3)
where v2 is the velocity of the skier at the top of the second hill.
Since the total mechanical energy is conserved, we equate the initial potential energy to the final sum of potential and kinetic energy:
P.E.1 = P.E.2 + K.E.2
m * g * h1 = m * g * h2 + (1/2) * m * v2^2
Canceling the mass (m) from both sides of the equation, we get:
g * h1 = g * h2 + (1/2) * v2^2
Rearranging the equation to solve for v2, we have:
v2^2 = 2 * g * (h1 - h2)
Taking the square root of both sides, we find:
v2 = √(2 * g * (h1 - h2))
Now, substitute the given values of h1, h2, and the acceleration due to gravity (g = 9.8 m/s^2) to find the velocity at the top of the second hill (v2).
b) To find the velocity of the skier at the bottom of the hill, we can use the conservation of energy principle again.
At the bottom of the hill, all of the potential energy is converted into kinetic energy:
P.E. bottom = 0 ...(4) (since the skier is at the bottom of the hill)
K.E. bottom = (1/2) * m * vbottom^2 ...(5) (where vbottom is the velocity at the bottom of the hill)
Using the conservation of energy principle:
Initial potential energy: P.E.1 = m * g * h1
Initial kinetic energy: K.E.1 = (1/2) * m * v1^2 ...(6) (where v1 is the velocity at the top of the first hill)
Since the total mechanical energy is conserved, we equate the initial potential energy to the final kinetic energy:
P.E.1 = K.E. bottom
m * g * h1 = (1/2) * m * vbottom^2
Canceling the mass (m) from both sides of the equation, we get:
g * h1 = (1/2) * vbottom^2
Rearranging the equation to solve for vbottom, we have:
vbottom^2 = 2 * g * h1
Taking the square root of both sides, we find:
vbottom = √(2 * g * h1)
Substitute the given value of h1 and the acceleration due to gravity to find the velocity at the bottom of the hill (vbottom).