A) \[{{v}_{B}}>{{v}_{A}}\]
B) \[{{v}_{A}}={{v}_{B}}\]
C) \[{{v}_{A}}>{{v}_{B}}\]
D) their velocities depends on their masses
Correct Answer: B
Solution :
From conservation of energy, potential energy at height h =K.E. at ground Therefore, at height h, P.E. of ball A \[P.E.={{m}_{A}}gh\] K.E. 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, \[{{v}_{A}}={{v}_{B}}\]You need to login to perform this action.
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