Question

The average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity. The velocity of the body as it reaches the ground is numerically equal to:

• A. $g$
• B. $\frac{g}{2}$
• C. $\frac{g}{\sqrt{2}}$
• D. $\sqrt{2}g$

Answer 1

The correct option is A $g$

We have average velocity $=\frac{g}{2}$
Since it is a free fall, average velocity, $v=\frac{S}{t}=\frac{\frac{1}{2}g{t}^2}{t}=\frac{g}{2}$, where $t$ is the time of descent and initial velocity is zero.

We know that for a free fall, velocity with which it reaches the ground: $V=gt$, where $t$ is time of descent .

Substituting the same in the above equation for average velocity:
$\frac{V}{2}=\frac{g}{2}$
$\therefore V=g$