A body in free fall takes eleven seconds to touch the ground. In the first second, it travels $4.9 \mathrm{~m}$, and in each subsequent second, the distance traveled increases by 9.8 m. What is the height of the fall and how many meters did it travel in the last second?
In the first second, the body travels $4.9 \, \mathrm{m}$, and since the distance traveled increases by $9.8 \, \mathrm{m}$ each second compared to the previous second, the body travels $4.9 + 9.8 \, \mathrm{m}$ in the second second, $4.9 + 2 \times 9.8 \, \mathrm{m}$ in the third second, $4.9 + 3 \times 9.8 \, \mathrm{m}$ in the fourth second, and so on, until the eleventh second, when the body travels $4.9 + 10 \times 9.8 = 102.9 \, \mathrm{m}$. The total distance traveled by the body until impact is

$$
\begin{aligned}
& 4.9 + (4.9 + 9.8) + (4.9 + 2 \times 9.8) + \cdots + (4.9 + 10 \times 9.8) \\
& \quad = 4.9 \times 11 + 9.8(1 + 2 + \cdots + 10) = 53.9 + 9.8 \times 55 = 592.9 \, \mathrm{m}
\end{aligned}
$$