Para hallar su energía cinética recordemos que:
[tex]\boxed{\boldsymbol{\mathsf{E_c=\dfrac{1}{2}m\cdot v^2}}} \hspace{30pt} \mathsf{Donde} \hspace{20pt}\begin{array}{ll}\mathsf{\blue{\rightarrow}\ E_c:Energ\acute{i}a\ cin\acute{e}tica(J)}\\\\\mathsf{\blue{\rightarrow}\ m:masa(kg)}\\\\\mathsf{\blue{\rightarrow}\ v:Rapidez(m/s)}\end{array}[/tex]
Extraigamos los datos del problema
[tex]\begin{array}{cccccccccccc} \mathsf{\bigcirc \kern -9pt \triangleright \:\:\:v = 25\:m/s}&&&&&&&&&&\mathsf{\bigcirc \kern -9pt \triangleright \:\:\:m = 145\ g=0.145\:kg}\end{array}[/tex]
Reemplacemos estos valores en nuestra ecuación escalar
[tex]\begin{array}{c}\mathsf{E_c=\dfrac{1}{2}m\cdot v^2}\\\\\\\mathsf{E_c=\dfrac{1}{2}(0.145\:kg)\cdot (25\:m/s)^2}\\\\\\\mathsf{E_c=\dfrac{1}{2}(0.145\:kg)(625\:m^2/s^2)}\\\\\\\mathsf{\boxed{\boxed{\boldsymbol{\mathsf{E_c=45.313\:J}}}}}\end{array}[/tex]
Rpta. La energía cinética de la pelota de béisbol es 45.313 Joules.
[tex]\boxed{\sf{{R}}\quad\raisebox{10pt}{$\sf{\red{O}}$}\!\!\!\!\raisebox{-10pt}{$\sf{\red{O}}$}\quad\raisebox{15pt}{$\sf{{G}}$}\!\!\!\!\raisebox{-15pt}{$\sf{{G}}$}\quad\raisebox{15pt}{$\sf{\red{H}}$}\!\!\!\!\raisebox{-15pt}{$\sf{\red{H}}$}\quad\raisebox{10pt}{$\sf{{E}}$}\!\!\!\!\raisebox{-10pt}{$\sf{{E}}$}\quad\sf{\red{R}}}\hspace{-64.5pt}\rule{10pt}{.2ex}\:\rule{3pt}{1ex}\rule{3pt}{1.5ex}\rule{3pt}{2ex}\rule{3pt}{1.5ex}\rule{3pt}{1ex}\:\rule{10pt}{.2ex}[/tex]
