A body is thrown up with a velocity 29.23 ms^-1 distance travelled in last second of upward motion is

Distance travelled in last n seconds of ascent incase of a body projected vertically is same as  the distance travelled by a freely falling body in first n seconds. 

$\therefore \mathrm{S}=\frac{1}{2}\mathrm{g}{\left(1\right)}^2=\frac{\mathrm{g}}{2}=\frac{9.8}{2}=4.9\mathrm{m}$