Question

A ball is allowed to fall from top of a building. If t, is
time taken to fall first 1/4 th of its height and t, is time
taken to fall last 1/4th
of its height then,t2/t1​

Answer 1

$\huge{\underline{\underline{.........Answer.........}}}$

$\huge{\underline{Given:-}}$

First 1/4 height by time ${t_1}$

Last 1/4 height by time ${t_2}$

Let the total height be H.

$\huge{\underline{Explanation:-}}$

Ball is dropped so,

u = 0.

the second kinematic equation becomes.

$s = \frac{1}{2} g {t}^{2}$

$\frac{H}{4} = \frac{1}{2} g {t_1}^{2} \:$

$t_1 = \sqrt{ \frac{H}{2g} }$

Now, the body takes time T to reach the ground.

$T = \sqrt{ \frac{2H}{g} }$

say,to cover 3H/4 height it takes time t'.

$\frac{1}{2} g {t'}^{2} = \frac{3H}{4}$

$t' = \sqrt{ \frac{3H}{2g} }$

Now,

${t_2}$ = T - ${t'}$

${t_2} = \sqrt{ \frac{2H}{g} - \frac{3H}{2g} }$

solving it.

$t_2 = \sqrt{ \frac{H}{g} } ( \sqrt{2} - \frac{ \sqrt{3} }{ \sqrt{2} } )$

$t_2 = \sqrt{ \frac{H}{g} } ( \frac{ \ \: 2 - \sqrt{3} }{ \sqrt{2} } ) \:$

Now,

$\frac{t_2 }{t_1} = \frac{ \frac{2 - \sqrt{3} }{ \sqrt{2}} }{ \sqrt{ \frac{1}{2} } }$

$\frac{t_2 }{t_1} = \frac{2 - \sqrt{3} }{1}$

$\frac{t_2 }{t_1} = 2 - \sqrt{3}$

$\boxed{\boxed{\frac{t_2 }{t_1} = 2 - \sqrt{3}}}$

So,the ratio is 2 - √3