Write the equations of motion when an object falling freely towards earth or thrown vertically upwards.

1. There are three equations of motion.
2. For free-falling bodies when falling with uniform accelerated motion, these equations of motion under uniform acceleration can be applied to the motion of freely falling bodies.
3. For a falling object, the acceleration due to gravity is ‘g’, so ‘a’ is replaced with ‘g’ while the distance ‘s’ of the freely falling bodies is replaced by the height ‘h’ of the freely falling bodies.

(a) The first equation of motion

$v=u+at$

becomes $v=u+gt$

$v=$final velocity

$u=$initial velocity

$g=$acceleration due to gravity

$t=$time taken by the body

(b) The second equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

becomes $h=ut+\frac{1}{2}g{t}^2$

$h=$ distance travelled by the body

$t=$ time taken

$u=$initial velocity

$g=$acceleration due to gravity

(c)the third equation of motion

$$\begin{gathered} v^2=u^2+2asbecomes \\ {v}^2=u^2+2gh \end{gathered}$$

$v=$ final velocity

$u=$initial velocity

$g=$ acceleration due to gravity

$h=$distance traveled by the body

Now,

(i) Equations of motion when an object falling freely towards earth. Thus acceleration of freely falling body a=g=9.8 m/s2 (Taking downward direction to be positive)

The initial speed of that body $\mathbf{}\mathit{u}\mathbf{=}\mathbf{0}$

Equations of motion :

$$\begin{gathered} \mathit{v}\mathbf{=}\mathit{g}\mathit{t} \\ \mathit{h}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{g}{\mathit{t}}^{\mathbf{2}} \\ {\mathit{v}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathit{g}\mathit{h} \end{gathered}$$

Equations of motion when an object thrown vertically upwards.

At the highest point final velocity becomes zero and acceleration due to gravity becomes negative.

$$\begin{gathered} \mathbf{}\mathbf{0}\mathbf{=}\mathit{u}\mathbf{-}\mathit{g}\mathit{t} \\ \mathit{u}\mathbf{=}\mathit{g}\mathit{t} \end{gathered}$$

Second equation becomes $\mathit{h}\mathbf{=}\mathit{u}\mathit{t}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{g}{\mathit{t}}^{\mathbf{2}}$

Equation of maximum height

$$\begin{gathered} \mathbf{0}\mathbf{=}{\mathit{u}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{g}\mathit{h} \\ \mathit{h}\mathbf{=}\frac{{\mathbf{u}}^{\mathbf{2}}}{\mathbf{2}\mathbf{g}} \end{gathered}$$