Answer An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by $g$. If atr resistance is neglected, the object is said to be in free fall. If the height through which the object falls is small compared to the earth's radius, $g$ can be taken to be constant, equal to $9.8 \mathrm{m} \mathrm{s}^{-2} .$ Free fall is thus a case of motion with uniform acceleration.

We assume that the motion is in $y$ -direction, more correctly in $-y$ -direction because we choose upward direction as positive. Since the acceleration due to gravity is always downward, it is in the negative direction and we have $a=-g=-9.8 \mathrm{m} \mathrm{s}^{-2}$

The object is released from rest at $y=0 .$ Therefore, $v_{0}=0$ and the equations of motion become:

$v=0-g t \quad=-9.8 t \quad \mathrm{m} \mathrm{s}^{-1}$

$y=0-1 / 2 g t^{2}=-4.9 t^{2} \quad \mathrm{m}$

$v^{2}=0-2 g y \quad=-19.6 y \quad \mathrm{m}^{2} \mathrm{s}^{-2}$

These equations give the velocity and the distance travelled as a function of time and also the vartation of velocity with distance. The variation of acceleration, velocity, and distance, with time have been plotted in below figures