Question

A body freely falling from rest has a velocity v after it falls through distance h.The distance it has to fall down further for its velocity to becomes double is?

Answer 1

For first 2 m,

$v^2=0^2+2g\times 2$

$v^2=4g$

After that let it traverse d m futher more when its v becomes double .i.e. 2v

$(2v{)}^2=v^2+2gd$

$2gd=3v^2$

$d=3v^2/2g$

put the value of $v^2$

d = $3\times 4g/2g=12/2=6meter.$