A) \[{{v}_{B}}>{{v}_{A}}\]
B) \[{{v}_{A}}={{v}_{B}}\]
C) \[{{v}_{A}}>{{v}_{B}}\]
D) their velocities depend on their masses
Correct Answer: B
Solution :
As there is no external force acting on the body any all the internal forces are conservative. So, we can apply the principle of conservation of energy. From conservation of energy, Potential energy at height h = KE at ground [\[\because \]KE at height = PE at ground = 0] Therefore, at height h, PE of ball A \[PE={{m}_{A}}gh\] KE at ground \[=\frac{1}{2}{{m}_{A}}v_{A}^{2}\] So, \[{{m}_{A}}gh=\frac{1}{2}{{m}_{A}}v_{A}^{2}\] \[{{v}_{A}}=\sqrt{2gh}\] Similarly, \[{{v}_{B}}=\sqrt{2gh}\] Therefore,\[{{v}_{A}}={{v}_{B}}\] NOTE In question, it is not mentioned that magnitude of thrown velocity of both balls are same which is assumed in solution.
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