142. A body, initially at rest, falls with a constant acceleration $g$. Find the law of motion.
Solution. $1^{0}$. It is known that the acceleration of a body moving in a straight line is the derivative of the velocity $v$ with respect to time $t$. Therefore, $g=\frac{d v}{d t}$, i.e., $d v=g d t$. From this, after integration, we find

$$
v=\int g d t=g \int d t=g t+C
$$

$2^{0}$. Using the initial conditions $v=0$ when $t=0$, we find $C_{1}=0$.

$3^{\circ}$. Thus, the equation of the velocity of motion has the form $v=g t$.

$4^{0}$. Now, let's use the fact that the velocity of the body is the derivative of the path $s$ with respect to time $t$, i.e., $v=\frac{d s}{d t}$. Since $v=g t$, then $\frac{d s}{d t}=g t$ or $d s=g t d t$. Integrating, we find

$$
\int d s=\int g t d t, s=\frac{g t^{2}}{2}+C_{2}
$$

$5^{0}$. Using the initial conditions $s=0$ when $t=0$, we get $C_{2}=0$.

$6^{\circ}$. Thus, the equation of motion of a freely falling body has the form $s=\frac{g t^{2}}{2}$.