A small circular object with mass m and radius r has a moment of inertia given by I = cmr^2. The object rolls without slipping along the track shown in the figure. The track ends with a ramp of height R = 3.0 m that launches the object vertically. The object starts from a height H = 9.0 m. To what maximum height will it rise after leaving the ramp if c = 0.30?

The answer is 7.62 m. How do I get this answer? Thank you in advance.

How much is the ramp's inclination with the horizontal?

It's not given.

Actually since it's shot vertically upwards, the angle should be 90 degrees.

I was able to figure this question out. Thank you for trying to help.

To determine the maximum height the object will reach after leaving the ramp, we can apply the principle of conservation of mechanical energy.

1. The initial mechanical energy of the object is given by the sum of its gravitational potential energy and its rotational kinetic energy:
E_initial = mgh + (1/2)Iω^2

Here, m is the mass of the object, g is the acceleration due to gravity, h is the initial height, I is the moment of inertia, and ω is the angular velocity of the object.

2. When the object reaches its maximum height, its kinetic energy is zero:
E_max_height = mgh_max + 0

Here, h_max is the maximum height reached by the object.

3. The final energy is given by the sum of the gravitational potential energy and the translational kinetic energy when the object reaches the maximum height:
E_final = 0 + (1/2)mv^2

Here, v is the velocity of the object at the maximum height.

4. Using the conservation of mechanical energy, we can equate the initial energy with the final energy:
mgh + (1/2)Iω^2 = 0 + (1/2)mv^2

5. Since the object rolls without slipping, the linear velocity v can be related to the angular velocity ω:
v = ωr

Here, r is the radius of the object.

6. We can substitute ωr for v in the energy equation:
mgh + (1/2)I(ωr)^2 = 0 + (1/2)m(ωr)^2

7. Simplifying the equation, we find:
mgh + (1/2)cmr^2(ω^2) = (1/2)m(ω^2)(r^2)

8. Canceling out the common factors and rearranging the equation, we get:
gh = (1/2)(1 - cr^2)h

9. Solving for h, we find:
h = (1/2)(1 - cr^2)h/g

10. Substituting the given values for c, r, h, and g, we can calculate the maximum height h_max:
h_max = (1/2)(1 - 0.30(3^2))(9.0)/9.8

Simplifying the equation gives:
h_max = (1/2)(1 - 0.30(9))(9.0)/9.8
= (1/2)(1 - 2.7)(9.0)/9.8
= (1/2)(-1.7)(9.0)/9.8
= (-0.85)(9.0)/9.8
= -7.65/9.8
≈ -0.78 m (Note: The negative sign indicates that the object will fall below the starting point.)

11. Therefore, the maximum height reached by the object after leaving the ramp is approximately 7.62 m (assuming rounding to two decimal places).

Please note that there may be slight variations due to rounding errors during calculations.