$(a)$ Two bodies have equal masses and are moving with uniform velocities of $2\,v$ and $3\,v$. What is the ratio of their kinetic energies?
$(b)$ $A$ mass is dropped from height $h$. At halfway to the ground,what is the total energy?

(A) Given: $m_1 = m, m_2 = m, v_1 = 2v, v_2 = 3v$. We need to find the ratio $E_1 / E_2$.
We know that kinetic energy $E = \frac{1}{2}mv^2$. Therefore:
$\frac{E_1}{E_2} = \frac{\frac{1}{2}m_1v_1^2}{\frac{1}{2}m_2v_2^2} = \frac{v_1^2}{v_2^2} = \frac{(2v)^2}{(3v)^2} = \frac{4v^2}{9v^2} = \frac{4}{9}$.
$(b)$ Total energy = Kinetic Energy $(KE)$ + Potential Energy $(PE)$.
At height $h$,the total energy is $mgh$ (since $KE = 0$ and $PE = mgh$).
According to the law of conservation of energy,the total energy remains constant at every point during the fall.
At halfway to the ground (height $h/2$),the velocity $v$ is given by $v^2 - u^2 = 2g(h/2)$. Since $u = 0$,$v^2 = gh$.
$KE = \frac{1}{2}mv^2 = \frac{1}{2}mgh$.
$PE = mg(h/2) = \frac{1}{2}mgh$.
Total energy = $KE + PE = \frac{1}{2}mgh + \frac{1}{2}mgh = mgh$.