# Question #cc79c

Nov 1, 2016

A falling object experiences two forces

1. Force due to gravity.
A downward force directed towards center of earth on the mass $m$.
${F}_{g} = m g$ .....(1)
where $g$ is acceleration due to gravity.
2. Drag force due to air resistance.
A force acting opposite to the direction of motion of the falling object. As such it is an upwards force slowing down the falling body.
Drag force ${F}_{D}$ depends on the the size, shape, and speed of the object. It also depends on physical properties of air and is given by the expression
${F}_{D} = \setminus \frac{1}{2} \setminus \rho \setminus {v}^{2} {C}_{D} \setminus A$ .....(2)
where $\rho$ is the density of air, $v$ is velocity of the object relative to air, $A$ is the area of cross section and ${C}_{D}$ is the drag coefficient which is a dimensionless number.

Therefore, net force ${F}_{\text{net}}$ acting on the falling body is
${F}_{\text{net}} = {F}_{g} - {F}_{D}$
Using (1) and (2) we get
${F}_{\text{net}} = m g - \setminus \frac{1}{2} \setminus \rho \setminus {v}^{2} {C}_{D} \setminus A$

Using equation for Newton's second law of motion we obtain
${a}_{\text{net"=F_"net}} / m$
$\implies {a}_{\text{net}} = \frac{m g - \setminus \frac{1}{2} \setminus \rho \setminus {v}^{2} {C}_{D} \setminus A}{m}$
$\implies {a}_{\text{net"=g-\frac {1}{2m}\rho \v^{2}C_{D}\A" ms}}^{-} 2$