Question

5.
A ball is allowed to fall from top of a building. If t1, is
time taken to fall first 1/4-th of its height and t2 is time
taken to fall last-1/4
th of its height then,
t1/t2, is​

Answer 1

Let height of building is h.

initial velocity of ball, u = 0

time taken to fall h/4 height from the top of building, t1 = √{2(h/4)/g} = √{h/2g}

now, find velocity at height = 3h/4

using formula, v² = u² + 2as

⇒v² = 0 + 2(g)(3h/4)

⇒v² = 3gh/2

⇒v = √{3gh/2} = √(1.5gh)

and velocity before striking the ground is v' = √{2gh}

now, time taken to fall last 1/4th of its height;

using formula, v = u + at

here, v = v' = √(2gh) , u = v = √(1.5gh)

a = -g and t = t2

so, √(1.5gh) = √(2gh) - gt2

t2 = √(2gh) - √(1.5gh)/g

= √(2h/g) - √(1.5h/g)

so, t1/t2 = √(1/2)/(√2 - √1.5)

= 1/(2 - √3)

hence, ratio of t1 and t2 = 1 : (2 - √3)

Answer 2

$\bf{\underline{\underline{SOLUTION :}}}$

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,

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

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

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

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