jmc

The cannonball is above $6$ meters in height when $-4.9t^2+14t-0.4\ge 6.$ Rearranging and multiplying through by $-10$, it follows that $$49t^2-140t+64\le 0.$$The quadratic expression factors as $$(7t-4)(7t-16)\le 0;$$then $7t-4,7t-16$ have opposite signs, so it follows that $\frac{4}{7}\le t\le \frac{16}{7}$. The cannonball then spends $\frac{16}{7}-\frac{4}{7}=\boxed{\frac{12}{7}}$ seconds above the height of $6$ meters.