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 body's 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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