Step 1: Find the initial velocity of the body ($u$). We are given the acceleration ($a$), final velocity ($v$), and time ($t$). We can use the first equation of motion: $$v = u + at$$ Given: $a = 12\,m/s^2$ $v = 80\,m/s$ $t = 5\,s$ Substitute the values into the equation: $$80\,m/s = u + (12\,m/s^2)(5\,s)$$ $$80 = u + 60$$ Subtract 60 from both sides to solve for $u$: $$u = 80 - 60$$ $$u = 20\,m/s$$ The initial velocity of the body is $\boxed{\text{20 m/s}}$. Step 2: Find the distance covered in the 9th second. The distance covered in the $n^{th}$ second is given by the formula: $$S_n = u + \frac{a}{2}(2n - 1)$$ Here, $n = 9$ (for the 9th second). We use the initial velocity $u = 20\,m/s$ (found in Step 1) and the acceleration $a = 12\,m/s^2$. Substitute the values into the formula: $$S_9 = 20\,m/s + \frac{12\,m/s^2}{2}(2 \times 9 - 1)$$ $$S_9 = 20 + 6(18 - 1)$$ $$S_9 = 20 + 6(17)$$ $$S_9 = 20 + 102$$ $$S_9 = 122\,m$$ The distance covered in the 9th second is $\boxed{\text{122 m}}$.