A stone is dropped from the top of a tower of height ‘H’ ., in last second of its journey it covers a distance of  0.36 H. Then height H and time of fall  is $\left(g=10m/s^2\right)$

Let ‘n’ is last sec $S_n$ is distance in last sec
Then from  $S=Ut+\frac{1}{2}a{t}^2,H=0+\frac{1}{2}g{n}^2--\left(1\right)$
Distance travelled before the last second is  $H-0.36H=0.64H$
 $$\begin{gathered} 0.64H=\frac{1}{2}g{\left(n-1\right)}^2 \\ substitute H value from equation \left(1\right) \\ \Rightarrow 0.64\left(\frac{1}{2}g{n}^2\right)=\frac{1}{2}g{\left(n-1\right)}^2 \end{gathered}$$
$\begin{array}{l}\Rightarrow 0.64n^2={\left(n-1\right)}^2\\ \Rightarrow \frac{64}{100}{n}^2={\left(n-1\right)}^2\\ \Rightarrow 8n=10n-10\\ \Rightarrow 2n=10\\ \Rightarrow n=5\end{array}$ 
The time of fall is 5sec
  $$\begin{gathered} H=\frac{1}{2}g{n}^2 \\ \Rightarrow H=\frac{1}{2}\left(10\right)\left(5^2\right) \\ \Rightarrow H=\frac{250}{2} \\ H=125m=height of the tower \end{gathered}$$