A body dropped from top of a tower fall through $$40\,m$$  during the last two seconds of its fall. The height of tower is
$$\left( {g = 10\,m/{s^2}} \right)$$

Solution :
Let the body falls through the height of tower in $$t$$ seconds.
From $${s_n} = u + \frac{a}{2}\left( {2n - 1} \right),$$
we have
Total distance travelled in last $$2\,s$$ of fall is
$$\eqalign{ & s = {s_t} + {s_{\left( {t - 1} \right)}} \cr & = \left[ {0 + \frac{g}{2}\left( {2t - 1} \right)} \right] + \left[ {0 + \frac{g}{2}\left( {2\left( {t - 1} \right) - 1} \right)} \right] \cr & = \frac{g}{2}\left( {2t - 1} \right) + \frac{g}{2}\left( {2t - 3} \right) \cr & = \frac{g}{2}\left( {4t - 4} \right) = \frac{{10}}{2} \times 4\left( {t - 1} \right) \cr & {\text{or}}\,\,40 = 20\left( {t - 1} \right){\text{ }} \cr & {\text{or}}\,\,t = 2 + 1 = 3\,s \cr} $$
Distance travelled in $$t$$ sec is
$$\eqalign{ & s = ut + \frac{1}{2}a{t^2} \cr & = 0 + \frac{1}{2} \times 10 \times {3^2} = 45\,m \cr} $$