91Ó°ÊÓ

The translational kinetic energy is defined as the work required accelerating it from rest to a given velocity of a rigid body.

\(\begin{array}{l}{v_0} = 25\;{\rm{m/s}}\\h = 28\;{\rm{m}}\end{array}\)

Conservation of energy

\(\frac{1}{2}mv_0^2 + \frac{1}{2}I\omega _1^2 = mgh + \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega _f^2 \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Rolling without slipper means \(\omega \;{\rm{ = }}\;\frac{v}{r}\)

\(\begin{array}{c}\frac{1}{2}I{\omega ^2} = \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\left( {\frac{v}{r}} \right)^2}\\ = \frac{1}{5}m{v^2} \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)\end{array}\)

Substitute (2) in (1),

\(\begin{array}{c}\frac{1}{2}mv_0^2 + \frac{1}{5}mv_0^2 = mgh + \frac{1}{2}mv_2^2 = \frac{1}{5}mv_2^2\\\frac{7}{{10}}mv_1^2 = mgh + \frac{7}{{10}}mv_2^2\\v_2^2 = \left( {v_1^2 - \frac{{10}}{7}gh} \right)\;\;\;\;\;\;\;\;{\rm{where}}\;{v_1} = 25\;{\rm{m/s}}\\{v_2} = 15.26\;{\rm{m/s}}\end{array}\)

Determine the time in air,

\(\begin{array}{c}{v_{oy}} = 0\\ay = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\y - yo = 28\;{\rm{m/s}}\\{\rm{28 = }}{{\rm{v}}_{oy}} + \frac{1}{2}ay{t^2}\\{t^2} = 5.71\\t = 2.39\;{\rm{s}}\end{array}\)

x component distance from cliff

\(\begin{array}{c}d = {v_f}t\\ = 15.26\;{\rm{m/s}} \times 2.39\;{\rm{s}}\\ = 36.47\;{\rm{m}}\\ \simeq 36.5\;{\rm{m}}\end{array}\)

Hence, the ball will go 36.5 m.

\(\begin{array}{c}{v_y} = {v_{oy}} + ayt\\ = 9.8 \times 2.39\\ = 23.42\;{\rm{m/s}}\\{{\rm{v}}_x} = \sqrt {\left( {{{\left( {23.42} \right)}^2} + {{\left( {15.26} \right)}^2}} \right)} \\ = 27.96\;{\rm{m/s}}\end{array}\)

Hence, the speed of the ball is 28 m/s.