Question

Find the ratio ofThe distance travelled by a freely falling body during the 50th second and 100th seconds

Answer 1

To find:

Ratio of distance travelled by a freely falling body during 50th and 100th second.

Calculation:

General formula for Displacement of a body during n th second is :

$\boxed{ \sf{s = u + \dfrac{1}{2} a(2n - 1)}}$

Here , u is initial velocity , a is acceleration and n refers to a particular second.

At 50th second :

$\sf{ \therefore \: s1 = 0 + \dfrac{1}{2} g(2n - 1)}$

$\sf{ = > \: s1 = 0 + \dfrac{1}{2} \times 10 \times \{(2 \times 50) - 1 \}}$

$\sf{ = > \: s1 = \dfrac{1}{2} \times 10 \times \{(2 \times 50) - 1 \}}$

$\sf{ = > \: s1 = 5\times \{(100) - 1 \}}$

$\sf{ = > \: s1 = 5\times 99}$

At 100th second:

$\sf{ \therefore \: s2= 0 + \dfrac{1}{2} g(2n - 1)}$

$\sf{ = > \: s2 = 0 + \dfrac{1}{2} \times 10 \times \{(2 \times 100) - 1 \}}$

$\sf{ = > \: s2 = 5\times \{(200) - 1 \}}$

$\sf{ = > \: s2 = 5\times 199}$

Hence required ratio:

$\rm{ \therefore \: \dfrac{s1}{s2} = \dfrac{5 \times 99}{5 \times 199} = \dfrac{99}{199} }$

$\rm{ = > s1 : s2 = 99 : 199 }$

Hence, final answer:

$\boxed{ \red{ \rm{ s1 : s2 = 99 : 199 }}}$