Two cannonballs, A and B, are dropped from rest from the same height h. Ball B has four times the mass of ball A. What is the ratio of their kinetic energies when they have fallen 3/4 of the distance?

Two cannonballs, A and B, are dropped from rest (initial velocity $v_0 = 0$) from a height $h$.
Ball B has four times the mass of ball A, so $m_B = 4m_A$. Let $m_A = m$, then $m_B = 4m$.
They fall under gravity. We can use the principle of conservation of mechanical energy, assuming no air resistance.
The total mechanical energy $E = K + U$ is conserved.
Initial energy $E_i = K_i + U_i = \frac{1}{2}mv_0^2 + mgh = 0 + mgh = mgh$.
Final energy $E_f = K_f + U_f = \frac{1}{2}mv_f^2 + mgy$.
Since energy is conserved, $E_i = E_f$.

$$mgh = \frac{1}{2}mv_f^2 + mgy$$

The balls are at a height $y = h/4$ above the ground. The distance fallen is $d = h - y = h - h/4 = \frac{3}{4}h$.
Using the kinematic equation $v_f^2 = v_0^2 + 2ad$, where $a=g$ and $v_0=0$:

$$v_f^2 = 0^2 + 2g\left(\frac{3}{4}h\right)$$ $$v_f^2 = \frac{3}{2}gh$$

The final velocity $v_f$ is the same for both balls because it only depends on the height fallen and $g$, not on mass.
So, $v_A = v_B = \sqrt{\frac{3}{2}gh}$.

The kinetic energy $K = \frac{1}{2}mv^2$.
For ball A:

$$K_A = \frac{1}{2}m_A v_A^2$$ $$K_A = \frac{1}{2}m \left(\frac{3}{2}gh\right)$$ $$K_A = \frac{3}{4}mgh$$

For ball B:

$$K_B = \frac{1}{2}m_B v_B^2$$ $$K_B = \frac{1}{2}(4m) \left(\frac{3}{2}gh\right)$$ $$K_B = 4 \times \frac{1}{2}m \left(\frac{3}{2}gh\right)$$ $$K_B = 4 K_A$$

From Step 3, we found $K_B = 4K_A$.

The final answer is $\boxed{\text{d. } K_B = 4K_A}$.