A body is projected with a velocity 60 ms^-1 vertically upwards the distance travelled in last second of its motion is $[\mathrm{g}=10{\mathrm{ms}}^{-1}]$

Distance traveled in the 1st second of the upward journey is equal to the distance traveled in the last second of the downward journey 
$\begin{array}{l}\mathrm{S}=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2\\ =60\left(1\right)-\frac{1}{2}\left(10\right)(1{)}^2=55\end{array}$