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Q.
A body falls from a height $h=200\, m$. The ratio of distance travelled in each $2\, s$, during $t=0$ to $t=6\, s$ of the journey is :
ManipalManipal 2006Motion in a Straight Line
Solution:
Using the relation
$s=u t+\frac{1}{2} g t^{2}$
As the body is falling from rest, $u=0$
$s=\frac{1}{2} g t^{2}$
Suppose the distance travelled in
$t=2 s, t=4 s, t=6\, s$
are $s_{2}, s_{4}$ and $s_{6}$ respectively
Now, $s_{2}=\frac{1}{2} g(2)^{2}=2\, g$
$s_{4}=\frac{1}{2} g(4)^{2}=8\, g$
$s_{6}=\frac{1}{2} g(6)^{2}=18\, g$
Hence, the distance travelled in first two seconds
$\left(s_{i}\right)_{2} =s_{2}-s_{0} =2\, g$
$\left(s_{m}\right)_{2}=s_{4}-s_{2} =8\, g-2\, g=6\, g$
$\left(s_{f}\right)_{2}=s_{6}-s_{4} =18\, g-8\, g=10\, g$
Now, the ratio becomes
$=2 g: 6 g: 10 g=1: 3: 5$