An object of uniform density rolls up a curved path with the initial velocity $$v_{\circ}$$ as shown in the figure. If the maximum height attained by an object is  $$\frac{7v_0^2}{10g}$$. (g = acceleration due to gravity), The object is a :

$$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2$$

$$K_{\text{total}} = mgh$$ (conservation of energy)

$$K_{\text{total}} = \frac{1}{2}mv_0^2 + \frac{1}{2}(kmR^2)\left(\frac{v_0}{R}\right)^2$$

$$K_{\text{total}} = \frac{1}{2}mv_0^2 + \frac{1}{2}kmv_0^2 = \frac{1}{2}mv_0^2(1 + k)$$

$$\frac{1}{2}mv_0^2(1 + k) = mgh$$

$$\frac{1}{2}v_0^2(1 + k) = gh \implies h = \frac{v_0^2(1 + k)}{2g}$$

$$\frac{v_0^2(1 + k)}{2g} = \frac{7v_0^2}{10g}$$

$$\frac{1 + k}{2} = \frac{7}{10}$$

$$k = 1.4 - 1 = 0.4 = \frac{2}{5}$$

Hence, it is a solid sphere.