Question

two stones of mass 100g and 200g fall from the same height what will be the ratio of time taken by both of them​

Answer 1

Explanation:

Explanation:$s = ut + frac{1}{2} {?at}^{2}$

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely falling

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²Becuse a=force /mass=f/m

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²Becuse a=force /mass=f/m S=1/2f/100*t²

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²Becuse a=force /mass=f/m S=1/2f/100*t²T=(200s/f)½

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²Becuse a=force /mass=f/m S=1/2f/100*t²T=(200s/f)½Body 2

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²Becuse a=force /mass=f/m S=1/2f/100*t²T=(200s/f)½Body 2S=1/2*F/200*t²

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²Becuse a=force /mass=f/m S=1/2f/100*t²T=(200s/f)½Body 2S=1/2*F/200*t²T=(400s/f)½this is 2nd time so ratio of the times

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²Becuse a=force /mass=f/m S=1/2f/100*t²T=(200s/f)½Body 2S=1/2*F/200*t²T=(400s/f)½this is 2nd time so ratio of the times T1/t2

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²Becuse a=force /mass=f/m S=1/2f/100*t²T=(200s/f)½Body 2S=1/2*F/200*t²T=(400s/f)½this is 2nd time so ratio of the times T1/t2[[{200s/f}]÷[{400s/f}]]½

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²Becuse a=force /mass=f/m S=1/2f/100*t²T=(200s/f)½Body 2S=1/2*F/200*t²T=(400s/f)½this is 2nd time so ratio of the times T1/t2[[{200s/f}]÷[{400s/f}]]½(1/2)½

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²Becuse a=force /mass=f/m S=1/2f/100*t²T=(200s/f)½Body 2S=1/2*F/200*t²T=(400s/f)½this is 2nd time so ratio of the times T1/t2[[{200s/f}]÷[{400s/f}]]½(1/2)½1/2½

Explanation:$s = ut + frac{1}{2} {?at}^{2}$As the u=0 S=1/2gt²A=g because its freely fallingFor 1st body S=1/2*F/m*t²Becuse a=force /mass=f/m S=1/2f/100*t²T=(200s/f)½Body 2S=1/2*F/200*t²T=(400s/f)½this is 2nd time so ratio of the times T1/t2[[{200s/f}]÷[{400s/f}]]½(1/2)½1/2½Mark as brainlliest answer and follow me