The apparent acceleration of the falling body is, $g’=g-a$, where $a$ is its vertical acceleration or real acceleration. So, when the vertical acceleration of the body changes, the apparent acceleration also changes. If the speed is constant, then acceleration $(a)$ is zero. Thus, $g’=g-0=g$. Hence, the acceleration of the falling body is simply the acceleration due to gravity $(g)$.
Home Gravitation Reasonings The acceleration of a falling body is measured inside an elevator travelling at constant speed 9.8 m/sec. What result is obtained?