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:
$g$
$\frac{g}{2}$
$\frac{g}{\sqrt{2}}$
$\sqrt{2} g$

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:
$g$
$\frac{g}{2}$
$\frac{g}{\sqrt{2}}$
$\sqrt{2} g$

Toppr Admin · Accepted Answer

We have average velocity -g2-xA0-Since it is a free fall- average velocity- v-St-12gt2t-g2-xA0-where t is the time of descent and initial velocity is zero-

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:gg2g√2√2g

We have average velocity =g2 Since it is a free fall, average velocity, v=St=12gt2t=g2, where t is the time of descent and initial velocity is zero.

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:
$g$
$\frac{g}{2}$
$\frac{g}{\sqrt{2}}$
$\sqrt{2} g$

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.