Question

A body is dropped from a certain height. Find the ratio of distances covered by it in 1st half of the

time of descent to the distance covered by it in the 2nd half of the time of descent​

Answer 1

CASE 1

Given:-

$u= 0\frac{m}{s}$

$a=10\frac{m}{s^{2} }$ (here 10 is used for easier calculations but use 9.8 at your own                 risk...…Haha! Just kiddin' let's move on!)

$S=\frac{h}{2}$

What to find:-

Time of Descent

Procedure:-

$s=ut+\frac{1}{2} at^{2}$

After plugging in the values, we get

$\frac{h}{2} =\frac{1}{2} 10t^{2}$

$h=10t^{2} \\\frac{h}{10} =t^{2} \\ t=\sqrt{\frac{h}{10} }$

CASE 2

Given:-

$v = 0\frac{m}{s}$

$a=10\frac{m}{s^{2} }$  (here same story as in case 1)

$S=\frac{h}{2}$

What to find:-

Time of Descent

Procedure:-

$v=u+aT$

After plugging in the values, we get

$0=u+10t$

$T=\frac{-u}{10}$

Both cases Comparision:-

$\frac{t}{T} =\frac{\sqrt{\frac{h}{20} } }{\frac{-u}{10} }$

$\sqrt{\frac{h}{20} } :-u/10$

Squaring on both sides,

$\frac{h}{20} :\frac{u^{2} }{100}$

$5h:u^{2}$