Question

The distance travelled by a body falling from rest in the first, second and third seconds are in the ratio ​

Answer 1

Answer:

2:4:9

Explanation:

when a body falls from rest

initial velocity,u=0m/s

We have the second equation of motion .i.e.,S=ut+1/2at^2

distance =S(1),S(2),S(3)

When  _ t=1

S(1)=0*t+1/2gt^2      (When a body is falling freely from a height ,acceleration                              

                                                    =acceleration due to gravity)

     =g/2

_t=2

S(2)=1/2 *g*2^2=2g

_t=3

S(3)=1/2*g*3^2=9/2g

Ratio=g/2:2g:9/2g

Multiplying by 2:

ratio=1:4;9