A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest of the second hill is circular, with a radius of 39.6 m. Neglect friction and air resistance. What must be the height h of the first hill so that the skier just loses contact with the snow at the crest of the second hill?

The speed V at the crest of the second hill must satisfy

V^2/R = g
where g = 9.8 m/s^2 and R = 39.6 m
V = 19.7 m/s

Use conservation of energy to get the required hill height H, measured above the crest of the second hill.

g*H = V^2/2

H = 19.8 m

Well, if we're neglecting friction and air resistance, we might as well neglect reality too! Let's imagine that instead of skis, the skier is actually riding a giant unicycle made out of marshmallows.

Now, let's put on our clown physics hats and solve this problem. To find the height h of the first hill, we need to consider the conservation of energy.

At the top of the first hill, the skier has potential energy, which is converted into kinetic energy as they slide down the hill. As the skier reaches the top of the second hill, they'll just lose contact with the snow.

At the crest of the second hill, all of the skier's potential energy will be converted into kinetic energy. So, we can equate the potential energy at the top of the first hill to the kinetic energy at the top of the second hill.

Now, let's throw in some numbers. The skier's mass doesn't matter because we're neglecting friction, so let's just assume they have the mass of a clown – about 50 kg.

At the top of the first hill, the skier will have potential energy given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height of the hill.

At the top of the second hill, the skier will have kinetic energy given by (1/2)mv^2, where v is the velocity of the skier at the crest.

Since energy is conserved, we can set mgh = (1/2)mv^2. We can cancel out the mass, and we'll be left with gh = (1/2)v^2.

Now, we need to relate the velocity v to the radius of the circular crest of the second hill. Since we're assuming no friction, the skier must just lose contact with the snow when the normal force becomes zero.

The normal force at the crest of the hill is given by N = mv^2/r, where r is the radius of the crest. Since the skier just loses contact with the snow, the normal force N must be zero. Hence, mv^2/r = 0.

Now, we've got ourselves into a bit of a pickle here. We have two equations: gh = (1/2)v^2 and mv^2/r = 0. Therefore, gh = 0. But gravity isn't zero, so h must be zero too!

So, to answer your question, in this magical world of marshmallow unicycles, the height h of the first hill must be zero for the skier to just lose contact with the snow at the crest of the second hill. Isn't physics fun?

To determine the height h of the first hill, let's consider the conservation of mechanical energy during the skier's motion. At the top of the first hill, the skier will have potential energy which will be converted into both kinetic energy and potential energy at the crest of the second hill.

The total mechanical energy of the skier can be calculated using the equation:

E_initial = E_final

At the top of the first hill (initial position), the skier only has potential energy, given by:

E_initial = m * g * h

where m is the mass of the skier, g is the acceleration due to gravity, and h is the height of the first hill.

At the crest of the second hill (final position), the skier will have both kinetic energy and potential energy, given by:

E_final = (1/2) * m * v^2 + m * g * R

where v is the velocity of the skier at the crest of the second hill and R is the radius of the circular crest.

Since the skier just loses contact with the snow at the crest of the second hill, we know that the normal force acting on the skier at this point is zero. This means that the gravitational force must equal the centripetal force:

m * g = m * v^2 / R

Simplifying the equation, we can express the velocity v as:

v = sqrt(g * R)

Substituting the values back into the equation for the final energy, we have:

E_final = (1/2) * m * (sqrt(g * R))^2 + m * g * R

Simplifying further:

E_final = (1/2) * m * g * R + m * g * R
E_final = (3/2) * m * g * R

Now we can set the initial energy equal to the final energy and solve for h:

m * g * h = (3/2) * m * g * R
h = (3/2) * R

Substituting the given value for the radius R = 39.6 m, we find:

h = (3/2) * 39.6 m
h ≈ 59.4 m

Therefore, the height h of the first hill must be approximately 59.4 m to allow the skier to just lose contact with the snow at the crest of the second hill.

To find the height h of the first hill, we can use the principle of conservation of energy.

At the top of the first hill, the skier has a certain amount of gravitational potential energy due to their height above the ground. As the skier descends the first hill and reaches the bottom, some of this potential energy is converted into kinetic energy.

When the skier reaches the bottom of the first hill, they have lost all their initial potential energy and have gained an equal amount of kinetic energy. This kinetic energy is then used to climb the second hill.

At the top of the second hill, the skier will have converted all their kinetic energy into potential energy. In order for the skier to just lose contact with the snow at the crest of the second hill, their potential energy at that point must be zero.

Using this information, we can equate the initial potential energy at the top of the first hill to the potential energy at the crest of the second hill:

mgh = 0

Where m is the mass of the skier, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the first hill.

Now, we need to find the relationship between the height of the second hill (H) and the radius of its crest (R) in order to relate it to the height of the first hill.

For a circular hill, the highest point is at the center of the circle, which is at a distance of R from the ground. Therefore, the height of the second hill can be calculated as:

H = R + R = 2R

So, the potential energy at the top of the second hill can be calculated as:

mgh = mg(2R)

Since the potential energy at the crest of the second hill must be zero, we can set this equation equal to zero:

mg(2R) = 0

Simplifying, we get:

2R = 0

Therefore, we have:

mgh = 0
mg(2R) = 0

Since the mass (m) and acceleration due to gravity (g) are common to both equations, we can divide them to eliminate the mass:

h / (2R) = 0 / 0

Since we have a division by zero, this equation is undefined.

Therefore, there is no specific height of the first hill at which the skier just loses contact with the snow at the crest of the second hill.