410. The height of a body thrown vertically upwards changes with time according to the law $H=200 t-4.9 t^{2}$. Find the velocity of the body at the end of the 10th second. How many seconds will the body fly upwards and what is the greatest height it will reach?
Solution. The velocity of the body is determined by the expression $v=\frac{d H}{d t}=$ $=200-9.8 t$; at $t=10$ we have $v=102($ m $/ \mathrm{s})$.

At the moment when the body reaches its maximum height, its velocity is zero. Therefore, for this moment $\frac{d H}{d t}=200-$ $-9.8 t=0$, from which $t=\frac{200}{9.8} \approx 20.4$ (s)

Substituting this value into the equation of motion, we get the maximum height to which the body rises:

$$
s=200 \cdot 20.4-4.9 \cdot 20.4^{2} \approx 2040 ; 8(\mathrm{m})
$$