Question

Body fall from rest velocity acquired by falling distance 2h

Answer 1

Suppose the body falls through a further distance of S then

Initial velocity(u)= 0

Wkt by equation of motion

v^2- u^2 =2gh

So v=(2gh)^(1/2)

v=sqrt(2gh)sqrt(2gh)

Now since the body falls a distance of S further

Now for this case u=sqrt(2gh)sqrt(2gh)

v=?

Applying the eq of motion again

Answer 2

Solution-

use equation v^2=u^2 +2gH

here u=0, H=2h

therefore, v^2= 0 +2 x g x 2h

v^2= 4gh

v=2(gh)^(1/2)