Question

A body falls from a height $h=200m$. The ratio of distance travelled in each $2s$, during $t=0$ to $t=6s$ of the journey is :

• A. $1:4:9$
• B. $1:2:4$
• C. $1:3:5$
• D. $1:2:3$

Answer 1

Answer: (C)

Using the relation
$s=ut+\frac{1}{2}g{t}^2$
As the body is falling from rest, $u=0$
$s=\frac{1}{2}g{t}^2$
Suppose the distance travelled in
$t=2s,t=4s,t=6s$
are $s_2,s_4$ and $s_6$ respectively
Now, $s_2=\frac{1}{2}g(2{)}^2=2g$
$s_4=\frac{1}{2}g(4{)}^2=8g$
$s_6=\frac{1}{2}g(6{)}^2=18g$
Hence, the distance travelled in first two seconds
${\left(s_i\right)}_2=s_2-s_0=2g$
${\left(s_m\right)}_2=s_4-s_2=8g-2g=6g$
${\left(s_f\right)}_2=s_6-s_4=18g-8g=10g$
Now, the ratio becomes
$=2g:6g:10g=1:3:5$