Question

If the distance travelled by a freely falling body in the last second of its journey is equal to the distance travelled in the first $2$ s, the time of descent of the body is

• A. $5s$
• B. $1.5s$
• C. $2.5s$
• D. $3s$

Answer 1

Answer: (C)

$2.5s$

Let total time is $u$ sec.
$s_{letsec}=s_n=\frac{1}{2}g{n}^2-\frac{1}{2}g(n-1{)}^2=\frac{1}{2}g(2n-1).....(1)$
$s_{travelled1st2}=s_2=\frac{1}{2}g(2{)}^2=2g.....(2)$
$\Rightarrow e{q}^n\left(1\right)=e{q}^n\left(2\right)$
$\Rightarrow \frac{1}{2}g(2n-1)=2g$

$\Rightarrow n=2.5sec$