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

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

Answer 1

Answer: (A)

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:
$\frac{V}{2}=\frac{g}{2}$
$\therefore V=g$