question_answer

A freely falling body has a velocity V after falling through a distance h. What is the distance it has to fall down further for its velocity to become 2V?

Answer:

In the first case,\[{{V}^{2}}={{0}^{2}}=2gh\Rightarrow {{V}^{2}}=2gh\,\,or\,\,2g=\frac{{{V}^{2}}}{h}\] In the second case, \[{{(2V)}^{2}}-{{V}^{2}}=2gs\] \[\Rightarrow 3{{V}^{2}}=\frac{{{V}^{2}}}{h}\times s\Rightarrow s=3h\]