question_answer

A body falling freely from rest has acquired a velocity v after it falls through a distance h. What is the distance it has to fall further for its velocity to become double?

Answer:

                Suppose body is freely falling from point A, where its initial velocity \[{{u}_{1}}=0.\] Let B be the point where its final velocity becomes \[{{V}_{1}}=v.\] Before coming to point B, it will cover a distance of \[{{S}_{1}}=h\] Acceleration of the body \[{{a}_{1}}=+g\] Let C be the point where its velocity \[({{v}^{2}})\]becomes 2v. Distance covered \[({{S}_{2}})\] before reaching point \[{{R}_{1}}=R\]

AB	BC
\[{{u}_{1}}=0\]	\[{{u}_{2}}=v\]
\[{{v}_{1}}=v\]	\[{{v}_{2}}=2v\]
\[{{S}_{1}}=h\]	\[{{S}_{2}}=x=?\]
\[{{a}_{1}}=+g\]	\[{{a}_{2}}=+g\]
\[({{v}^{2}})-({{0}^{2}})=2gh\]	\[{{(2v)}^{2}}-{{v}^{2}}=2gx\]

\[{{v}^{2}}=2gh\]             ?..(1)    \[3{{v}^{2}}=2gx\]           ?.(2) Substituting (1) in (2), we get, \[3\,2gh=2gx\] \[x=3h\]