A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity 3 m s$$^{-1}$$ (as shown in figure). Maximum height with respect to the initial position covered by it will be _______ cm.
(take, $$g = 10$$ m s$$^{-2}$$)

We have a hollow spherical ball rolling up a curved surface with initial velocity $$u = 3$$ m/s. For a hollow sphere, the moment of inertia is $$I = \frac{2}{3}mR^2$$.

Applying conservation of energy, the total kinetic energy at the bottom equals the potential energy at the maximum height:

$$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$$

Since the ball rolls without slipping, $$v = R\omega$$. Substituting:

$$\frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{3}mR^2\right)\frac{v^2}{R^2} = mgh$$

$$\frac{1}{2}mv^2 + \frac{1}{3}mv^2 = mgh$$

$$\frac{5}{6}mv^2 = mgh$$

Now, solving for $$h$$:

$$h = \frac{5v^2}{6g} = \frac{5 \times 9}{6 \times 10} = \frac{45}{60} = 0.75 \text{ m} = 75 \text{ cm}$$

So, the answer is $$75$$ cm.