Answer:[tex]v=\sqrt{\frac{10}{7}gH}[/tex]
Explanation:
Given
Marble rolls down the hill
Considering it to be a solid sphere of radius R
Conserving the total Energy as it is constant
[tex]E_0=E_f\quad [\text{as initial energy=Final energy}][/tex]
[tex]MgH=\frac{1}{2}Mv^2+\frac{1}{2}I\omega ^2[/tex]
where I=moment of inertia of marble
[tex](\frac{2}{5}MR^2)[/tex]
Considering pure rolling
so [tex]v=\omega \times R[/tex]
[tex]MgH=\frac{1}{2}Mv^2+\frac{1}{2}\times \frac{2}{5}MR^2\times (\frac{v}{R})^2[/tex]
[tex]gH=\frac{7}{10}v^2[/tex]
[tex]v=\sqrt{\frac{10}{7}gH}[/tex]
