Question

A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first $5$ seconds. It remained in air for:

• A. $12s$
• B. $13s$
• C. $25s$
• D. $26s$

Answer 1

The correct option is B $13s$

Let the time for which it remained in air be $t$.

Initial velocity of stone $u=0$ m/s

Distance covered in first 5 second, $S=ut+\frac{1}{2}g{t}^2$ where $t=5$ s

$\therefore$ $S=0+\frac{1}{2}\times 10\times 5^2=125$ m

Distance covered by stone in last second of motion $S^{\prime }=S_t-S_{t-1}$

OR $S^{\prime }=ut+\frac{1}{2}g{t}^2-\left[u\right(t-1)+\frac{1}{2}g(t-1{)}^2]$ where $u=0$

OR $S^{\prime }=\frac{1}{2}g[t^2-(t-1{)}^2]$

$⟹$ $S^{\prime }=\frac{1}{2}g(2t-1)$

But $S^{\prime }=S$

$\frac{1}{2}\times 10(2t-1)=125$

We get $2t-1=25$

$⟹$ $t=13s$