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

Consider a body which is projected vertically up with an initial velocity$=\text{u}$

$\text{a}=-g$

Let the time ascent$=\text{t}$

We know$\text{t}=\text{u/g}$  

Then the height ascended in last 1/2 second = ?

i.e. the height ascended in last 1/2s = height ascended in ${\text{t}}_{\text{ase}}-\text{height assended}{\text{(t}}_{\text{ase}}-\text{1/2)}$  $=\text{u}\left(\frac{\text{u}}{\text{g}}\right)-\frac{\text{1}}{\text{2}}\left(\text{g}\right){\left(\frac{\text{u}}{\text{g}}\right)}^{\text{2}}$ $-\left[\left(\text{u}\right)\left(\frac{\text{u}}{\text{g}}-\frac{\text{1}}{\text{2}}\right)+\frac{\text{1}}{\text{2}}\text{g}{\left(\frac{\text{4}}{\text{g}}-\frac{\text{1}}{\text{2}}\right)}^2\right]$

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

Height of ascended in last1/2second $=\frac{\text{g}}{\text{8}}$  

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