A) \[\frac{S}{4},\frac{3gS}{2}\]
B) \[\frac{S}{4},\frac{\sqrt{3gS}}{2}\]
C) \[\frac{S}{2},\frac{\sqrt{3gS}}{2}\]
D) \[\frac{S}{2},\frac{\sqrt{3gS}}{2}\]
E) \[\frac{S}{3},\frac{\sqrt{3gS}}{2}\]
Correct Answer: D
Solution :
We can realise the situation as shown. Let at point C distance\[x\]from highest point A, the particles kinetic energy is three times its potential energy. Velocity at C, \[{{v}^{2}}=0+2gx\] or \[{{v}^{2}}=2gx\] ...(i) Potential energy at C \[=mg(S-x)\] ...(ii) At point C, Kinetic energy\[=3\times \] potential energy ie, \[\frac{1}{2}m\times 2gx=3\times mg(S-x)\] or \[x=3S-3x\] or \[4x=3S\] or \[S=\frac{4}{3}x\] or \[x=\frac{3}{4}S\] Therefore, from Eq. (i) \[{{v}^{2}}=2g\times \frac{3}{4}S\] Or \[{{v}^{2}}=\frac{3}{2}gS\]or\[V=\sqrt{\frac{3}{2}gS}\] Height of the particle from the ground \[=S-x\] \[=S-\frac{3}{4}S=\frac{S}{4}\]You need to login to perform this action.
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