Question

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

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

Answer 1

The correct option is C $1:3:5$

Let $h_1,h_2$ and $h_3$ be the distances travelled in each $2$ seconds.
Using second equation of motion, we know
$S=ut+\frac{1}{2}a{t}^2$
As the body is at rest initially
$\Rightarrow h_1=\frac{1}{2}\times g\times (2{)}^2=2g$
$h_2$ = Distance covered in $4\text{s}$ $-$ Distance covered in $2\text{s}$
$=\frac{1}{2}\times g\times (4{)}^2-2g=6g$
$h_3$ = Distance covered in $6\text{s}-$Distance covered in $4\text{s}$
$=\frac{1}{2}\times g\times (6{)}^2-\frac{1}{2}\times g\times (4{)}^2=10g$
$\therefore h_1:h_2:h_3=1:3:5$