Question

A body falls from rest. In the last second of its fall n coven half of the total distance. If g is $9.8m/s^2$, then the total time of its fall is (in second)

• A. $2$
• B. $2+\sqrt{2}$
• C. $2-\sqrt{2}$
• D. $2\pm \sqrt{2}$

Answer 1

Answer: (D)

$2\pm \sqrt{2}$

Let the total time taken by the body to fall be $t$.

Initial velocity of the body $u=0m/s$

Total distance traveled $S=ut+\frac{1}{2}a{t}^2$

$\therefore$ $S=\frac{1}{2}a{t}^2$

Distance covered in last second, $S^{\prime }=u+\frac{1}{2}a(2t-1)$

$\therefore$ $S^{\prime }=\frac{1}{2}a(2t-1)$

But $S^{\prime }=\frac{1}{2}S$

$\therefore$ $\frac{1}{2}a(2t-1)=$ $\frac{1}{4}a{t}^2$

We get $t^2-4t+2=0$

$⟹$ $t=\frac{4\pm \sqrt{(-4{)}^2-4(1\left)\right(2)}}{2\left(1\right)}$

Or $t=\frac{4\pm \sqrt{8}}{2}$

Or $t=2\pm \sqrt{2}$