A ball is dropped from a height H from rest. The ball travels...

Given: Height of free fall of ball is& $H$

Distance covered by the ball is $\dfrac{H}{2}$~in last~$1 . 0s$.

Now, displacement in the $n^{th}$~second~{s}_{n} = u + a \left(n - \dfrac{1}{2}\right)</p><p>Here~{s}_{n} = $\dfrac{H}{2},~$u = 0$~(since the ball is at rest initially)</p><p>Thus,~\dfrac{H}{2} =$g \left(n - \dfrac{1}{2}\right) . . . \left(1\right)$</p><p>where~$n$~is the time taken to reach the ground~$n = $\sqrt{\dfrac{2H}{g}}</p><p>\dfrac{n^{2}g}{2} =$H$</p><p>Substituting the value of~$n$~in equation \left(1\right)$

\dfrac{n^{2}g}{2 \times 2} =g \left(n - \dfrac{1}{2}\right)</p><p>n^{2} = 4n - 2$</p><p>$n^{2} - 4n + 2 = 0$</p><p>$n = 2 $\pm \sqrt{2}</p><p>$n$~should be greater than~$1$</p><p>so,~$n = 2 + $\sqrt{2} =$3 . 41s$</p>$