Question

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

Answer 1

$\huge{\underline{\underline{.........Answer.........}}} </p><p>.........Answer.........$

$\huge{\underline{Given:-}} </p><p>Given:−$

First 1/4 height by time {t_1}t

1

Last 1/4 height by time {t_2}t

2

Let the total height be H.

$\huge{\underline{Explanation:-}} </p><p>Explanation:−$

Ball is dropped so,

u = 0.

the second kinematic equation becomes.

$s = \frac{1}{2} g {t}^{2}s= </p><p>2</p><p>1</p><p> </p><p> gt </p><p>2</p><p> </p><p></p><p>\frac{H}{4} = \frac{1}{2} g {t_1}^{2} \: </p><p>4</p><p>H</p><p> </p><p> = </p><p>2</p><p>1</p><p> </p><p> gt </p><p>1</p><p> </p><p> </p><p>2</p><p> </p><p></p><p>t_1 = \sqrt{ \frac{H}{2g} }t </p><p>1</p><p> </p><p> = </p><p>2g</p><p>H$

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

$T = \sqrt{ \frac{2H}{g} }T= </p><p>g</p><p>2H$

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

$\frac{1}{2} g {t'}^{2} = \frac{3H}{4} </p><p>2</p><p>1</p><p> </p><p> gt </p><p>′</p><p> </p><p>2</p><p> = </p><p>4</p><p>3H$

Now,

${t_2}t </p><p>2</p><p> </p><p> = T - {t'}t </p><p>′$

$</p><p>{t_2} = \sqrt{ \frac{2H}{g} - \frac{3H}{2g} }t </p><p>2</p><p> </p><p> = </p><p>g</p><p>2H</p><p> </p><p> − </p><p>2g</p><p>3H</p><p>$

solving it.

$t_2 = \sqrt{ \frac{H}{g} } ( \sqrt{2} - \frac{ \sqrt{3} }{ \sqrt{2} } )t </p><p>2$

$= </p><p>g</p><p>H</p><p> </p><p> </p><p> </p><p> ( </p><p>2</p><p> </p><p> − </p><p>2</p><p> </p><p> </p><p>3</p><p> </p><p> </p><p> </p><p> )</p><p></p><p>$

$t_2 = \sqrt{ \frac{H}{g} } ( \frac{ \ \: 2 - \sqrt{3} }{ \sqrt{2} } ) \:t </p><p>2</p><p>$

$= </p><p>g</p><p>H</p><p> </p><p> </p><p> </p><p> ( </p><p>2</p><p> </p><p> </p><p> 2− </p><p>3</p><p> </p><p> </p><p> </p><p> )$

Now,

$\frac{t_2 }{t_1} = \frac{ \frac{2 - \sqrt{3} }{ \sqrt{2}} }{ \sqrt{ \frac{1}{2} } } </p><p>t </p><p>1</p><p> </p><p> </p><p>t </p><p>2$

$</p><p> = </p><p>2</p><p>1</p><p> </p><p> </p><p> </p><p> </p><p>2</p><p> </p><p> </p><p>2− </p><p>3$

$\frac{t_2 }{t_1} = \frac{2 - \sqrt{3} }{1} </p><p>t </p><p>1</p><p> </p><p> </p><p>t </p><p>2$

$= </p><p>1</p><p>2− </p><p>3</p><p> </p><p> </p><p> </p><p></p><p>$

$\frac{t_2 }{t_1} = 2 - \sqrt{3} </p><p>t </p><p>1</p><p> </p><p> </p><p>t </p><p>2</p><p> </p><p> </p><p> </p><p> =2− </p><p>3$

$\boxed{\boxed{\frac{t_2 }{t_1} = 2 - \sqrt{3}}} </p><p>t </p><p>1</p><p> </p><p> </p><p>t </p><p>2</p><p> </p><p> </p><p> </p><p> =2− </p><p>3</p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><p>So,the ratio is 2 - √3</p><p></p><p> </p><p>$