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

Question

A body freely falling from the rest has a velocity v after his falls through a height h. The distance, it has to fall down further for its velocity becomes double, is (A)  4 text h  (B)  2 text s  (C)  1/√2s  (D)  √2s

Tardigrade Admin · Accepted Answer

Here, initial velocity of the body v1=v Initial height h=hc Final velocity of the body v2=2v Now, from the equation of motion v2=u2+2gh v2 propto h So, [ (v1/v2) ]2=[ (h1/h2) ] Or, [ (v/2v) ]2=(h/h2) Or, (h/h2)=(1/4) Or, h2=4h

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

$4 h$

$2 s$

$1/ 2 s$

$2 s$

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

4h

2s

1/2​s

2​s

Here, initial velocity of the body v1​=v Initial height h=hc Final velocity of the body v2=2v Now, from the equation of motion v2=u2+2gh v2∝h So, [v2​v1​​]2=[h2​h1​​] Or, [2vv​]2=h2​h​ Or, h2​h​=41​ Or, h2​=4h

$4 h$

$2 s$

$1/ 2 s$

$2 s$