Question

The distance travelled by a body falling freely from rest in 2nd 3rd and 5th second of its motion are in the ratio a)9:5:3. b)3:5:9. c)5:3:9. d)5:9:3. please answer it with process I want the process I know the answer but I need the process. please help me

Answer 1

Answer:

3:5:9.

Explanation:

the basic formula to solve this question is

= 2n - 1

now, put n = 2 then = 2*2-1 = 3

for n= 2

= 2*3-1= 5

now n= 3.

= 2* 5- 1 = 9

Answer 2

Answer:

The ratio of distance travelled in 2nd 3rd and 5th second is $3:5:9$

Explanation:

Given, the body is starting from rest,

                    therefore initial velocity, $u=0$

            the body is freely falling,

                    therefore acceleration, $a=g=10m/s^{2}$

           and the time, $t=2nd, \ 3rd \ \&\ 5th\ sec.$

Using the equation of motion, for distance travelled in nth second is

                                         $S_{n} =u+\frac{a}{2}(2n-1)$

For,  $u=0$ , and $a=g$,     $S_{n} =\frac{g}{2}(2n-1)$

As $g$ is same for all the cases,

                                        $S_{n} \propto (2n-1)$

Therefore, the distance travelled in 2nd second is,

                                  $S_{2} =(2\times 2-1)=3$

Distance travelled in 3rd second is, $S_{3} =(2\times 3-1)=5$

Distance travelled in 5th second is, $S_{5} =(2\times 5-1)=9$

Hence the ratio of distance travelled in 2nd 3rd and 5th second is $3:5:9$

so, the correct answer is option b.