A) \[\sqrt{mgh}\]
B) \[m\sqrt{2gh}\]
C) \[m\sqrt{gh}\]
D) zero
Correct Answer: B
Solution :
When the body is falling freely from height h, then from third equation of motion \[{{v}^{2}}-{{u}^{2}}=2gh\] \[(\because \,u=0)\] \[\therefore \] \[{{v}^{2}}=2gh\Rightarrow v=\sqrt{2gh}\] Now, linear momentum of body is given \[p=m\text{ }v\] From Eqs.(i) and (ii) Momentum \[=m\sqrt{2gh}\]
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