A) \[1:1\]
B) \[1:4\]
C) \[1:7\]
D) \[1:8\]
Correct Answer: B
Solution :
In first case, let at point\[P\], its kinetic and potential energies are equal\[ie,\] \[\frac{1}{2}m{{v}^{2}}=mgh\] \[\Rightarrow \] \[h=\frac{{{v}^{2}}}{2g}\] ? (i) In second case, when bodys velocity is\[2v\]then at the same point\[P\] \[\frac{PE}{KE}=\frac{mg\times \frac{{{v}^{2}}}{2g}}{\frac{1}{2}m{{(2v)}^{2}}}=\frac{1}{4}\]You need to login to perform this action.
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