A) \[\sqrt{3}\,{{\nu }_{0}}\]
B) \[3\,{{\nu }_{0}}\]
C) \[9\,{{\nu }_{0}}\]
D) \[3/2\,{{\nu }_{0}}\]
Correct Answer: A
Solution :
At maximum height velocity is zero. From equation of motion we have, \[{{v}^{2}}={{u}^{2}}-2gh\] where v is final velocity and u is initial velocity Since ball reaches maximum height, velocity at the highest point is zero. Therefore, we have \[\operatorname{v} = 0,\,\,u={{v}_{0}}\] \[0=v_{0}^{2}-2gh\] \[\Rightarrow \,\,\,\,\,\,{{v}_{0}}\,\,=\,\,\sqrt{2gh}\] when \[\operatorname{h}'=3h\], then \[v_{0}^{'}=\sqrt{2g\times 3h}\,\,=\,\,\sqrt{3}\,\,\sqrt{2gh}=\sqrt{3}\,{{\nu }_{0}}\]
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