371. A body is thrown vertically upward with a velocity that varies according to the law \( v = (29.4 - 9.8 t) \) m/s. Find the maximum height of ascent.
Solution. Let's find the time during which the body was rising: $29.4 - 9.8 t = 0$ (at the moment of maximum ascent, the velocity is zero); $t = 3$ (s). Therefore,

$$
s=\int_{0}^{3}(29.4 - 9.8 t) \, dt = 9.8 \left(3 t - \frac{t^{2}}{2}\right) \bigg|_{0}^{3} = 44.1 \quad (\mathrm{m})
$$