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A body is dropped from a certain height and takes 6 second to reach the ground.How much time does it take to cover one-ninth the distance starting from top?

Susmita
425 Points
2 years ago
$h=ut+ \frac{1}{2}gt^2$
initial speed u=0.
t=6s
acceleration g=9.81 m/s2=10m/s2 approx.
Therefore , height $h= \frac{1}{2}*10*6^2=5*36 m$
Now one nineth the distance  is
$H=\frac{h}{9}= \frac{1}{9}*5*36=5*4 m=20m$
Using the same formula
$t= \sqrt[]{2h/g}=\sqrt[]{\frac{2*20}{10}}=2s$