Explanation

Let the body fall through the height of tower in t seconds. From, $D_{n} = u + a/2(2n - 1)$ we have, total distance travelled in last 2 second of fall is $D = D_{t} + D_{(t-1)}$ = $[0 + g/2(2t - 1)] + [0 + g/2\{2(t - 1) - 1\}]$ = $10/2 \times 4(t - 1)$ or, $40 = 20(t - 1) + g/2(2t - 3) = g/2(4t - 4)$ Distance travelled in t second is $s = ut + \frac{1}{2}a t^{2} = 0 + \frac{1}{2} \times 10 \times 3^{2} = 45 m$