Question

A solid sphere of mass $2\, kg$ rolls up a $30^{\circ}$ incline with an initial speed of $10 \,m/s$. The maximum height reached by the sphere is $(g = 10\,m/s^2)$

• A. 3.5 m
• B. 7.0 m
• C. 10.5 m
• D. 14.0 m

Answer 1

Answer: (B)

7.0 m

Answer 2

$m g h =\frac{1}{2} m v^{2}\left(1+\frac{K^{2}}{R^{2}}\right)$
$\Rightarrow \frac{7}{10} m v^{2} =m g h$
$\left( \because \frac{K^2}{R^2} = \frac{2}{5} \right)$
$\Rightarrow h =\frac{7}{10}\left(\frac{v^{2}}{g}\right)$
Given, $v=10 \,m / s , g=10 \,m / s ^{2} .$
$\therefore h=\frac{7}{10} \times \frac{10 \times 10}{10}$
$\Rightarrow h=7 \,m$