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(B) Let the total height of the tower be $H$ and the total time of fall be $t$.Since the stone is dropped,the initial velocity $u = 0$.The total heigh

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$3$

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(B) Let the total height of the tower be $H$ and the total time of fall be $t$.Since the stone is dropped,the initial velocity $u = 0$.The total height $H = \frac{1}{2}gt^2$.The distance covered in the first $(t-1)$ seconds is $H' = \frac{1}{2}g(t-1)^2$.The distance covered in the last second is $H - H' = \frac{5}{9}H$.Substituting the expressions for $H$ and $H'$,we get:$\frac{1}{2}gt^2 - \frac{1}{2}g(t-1)^2 = \frac{5}{9} \left(\frac{1}{2}gt^2ight)$Dividing both sides by $\frac{1}{2}g$:$t^2 - (t-1)^2 = \frac{5}{9}t^2$$t^2 - (t^2 - 2t + 1) = \frac{5}{9}t^2$$2t - 1 = \frac{5}{9}t^2$$18t - 9 = 5t^2$$5t^2 - 18t + 9 = 0$Solving the quadratic equation using the factorization method:$5t^2 - 15t - 3t + 9 = 0$$5t(t - 3) - 3(t - 3) = 0$$(5t - 3)(t - 3) = 0$This gives $t = \frac{3}{5} 	ext{ s}$ or $t = 3 	ext{ s}$.Since the last second implies $t > 1$,we reject $t = \frac{3}{5} 	ext{ s}$.Therefore,the total time of fall is $3 	ext{ s}$.

$A$ stone dropped from the top of a tower is found to travel $\left(\frac{5}{9}\right)$ of the height of the tower during the last second of its fall. The time of fall is ....... $s$.

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