Question

As 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 is in air for :

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

Answer 1

Answer: (C)

Distance travelled in nth $\sec s_{n}=u+\frac{1}{2} g(2 n-1)$
In case of freely falling body $u =0$
So, $s_{n}=\frac{1}{2} g(2 n-1)$...(1)
Now distance travelled in $t\, \sec$, is
$s=\frac{1}{2} g \times(5)^{2}=\frac{25}{2} g$ ...(2)
When $t=5\, \sec$, then $s=\frac{1}{2} g \times(5)^{2}=\frac{25}{2} g$ ...(2)
According to the condition $s_{n}=s$,
so, from eqs. (1) and (2)
$\frac{1}{2} g(2 n-1)=\frac{g}{2} \times 25$
or $2 n-1 =25$ or $n=13\, \sec$