A) 68%
B) 41%
C) 200%
D) 100%
Correct Answer: B
Solution :
[b] Momentum \[P=mv=m\sqrt{2gh}\] \[\left( \therefore {{v}^{2}}={{u}^{2}}+2gh;\text{ Here }u=0 \right)\] When stone hits the ground momentum \[P=m\sqrt{2gh}\] When same stone dropped from 2h (100% of initial) then momentum \[{P}'=m\sqrt{2g(2h)}=\sqrt{2}P\] Which is changed by 41 % of initial.You need to login to perform this action.
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