The distance described during the last half second of upward motion, of a body projected vertically upward is

Solution:

Consider a body which is projected vertically up with an initial velocity = u

$a=-g$

Let the time ascent = t

$\mathrm{We}\text{ know }\mathrm{t}=\mathrm{u}/\mathrm{g}$

$\text{ Then the height ascended in last }1/2\text{ second }=\text{ ? }$

i.e. the height ascended in last 1/2s = height ascended in 

${\mathrm{t}}_{\mathrm{ass}}\text{-height assended }\left({\mathrm{t}}_{\mathrm{as}}-1/2\right)=\mathrm{u}\left(\frac{\mathrm{u}}{\mathrm{g}}\right)-\frac{1}{2}\left(\mathrm{g}\right){\left(\frac{\mathrm{u}}{\mathrm{g}}\right)}^2-\left[\left(\mathrm{u}\right)\left(\frac{\mathrm{u}}{\mathrm{g}}-\frac{1}{2}\right)+\frac{1}{2}\mathrm{g}{\left(\frac{4}{\mathrm{g}}-\frac{1}{2}\right)}^2\right]$

$\Rightarrow \frac{{\mathrm{u}}^2}{\mathrm{g}}-\frac{{\mathrm{u}}^2}{2\mathrm{g}}-\frac{{\mathrm{u}}^2}{\mathrm{g}}+\frac{\mathrm{u}}{2}-\frac{{\mathrm{u}}^2}{2\mathrm{g}}-\frac{\mathrm{u}}{2}+\frac{\mathrm{g}}{8}$

$\text{ Height of ascended in last }1/2\text{ second }=\frac{\mathrm{g}}{8}$

We get this value is independent of velocity of projection and time of ascent and mass of the body it is always constant.