Answer:
Let v be the velocity acquired by the body when it travels h metre starting from rest. The distance it has to fall down further for its velocity to become double is given as with final velocity v and initial velocity equal to zero with final velocity 2v and initial velocity v after travelling h metre. \[{{v}^{2}}{{o}^{2}}=2gh\] \[{{v}^{2}}=2gh\] .... (1) \[{{(2v)}^{2}}{{v}^{2}}=2g{{h}^{1}}\] \[3{{v}^{2}}=2g{{h}^{1}}\] ...(2) \[\frac{2}{4}=\frac{3{{v}^{2}}}{{{v}^{2}}}=\frac{2g{{h}^{1}}}{2gh}\]\[\Rightarrow \]\[{{h}^{1}}=3h\]
You need to login to perform this action.
You will be redirected in
3 sec