A) \[\sqrt{2gH}\]
B) \[\sqrt{gH}\]
C) \[\frac{\sqrt{gH}}{2}\]
D) \[\frac{\sqrt{2g}}{H}\]
Correct Answer: B
Solution :
[b] Suppose the two bodies A and B meet at time t, at a height from group for body B, u =0' \[h=H/2\] \[h=ut+\frac{1}{2}g{{t}^{2}}\left[ u=0,h=\frac{H}{2} \right]\] \[\frac{H}{2}=\frac{1}{2}g{{t}^{2}}......(1)\] For body \[A,u={{v}_{{}^\circ }},h=\frac{H}{2}\] \[h=ut-\frac{1}{2}g{{t}^{2}}\] \[h=ut-\frac{1}{2}g{{t}^{2}}......(2)\] From eq 1 and 2 \[{{V}_{{}^\circ }}t\frac{1}{2}g{{t}^{2}}=\frac{1}{2}g{{t}^{2}}\] \[t=\frac{{{V}_{{}^\circ }}}{g}\] Put this value in (1) \[\frac{H}{2}=\frac{1}{2}g\frac{{{V}_{{}^\circ }}}{{{g}^{2}}}\Rightarrow H=\frac{V_{{}^\circ }^{2}}{g}\] \[{{V}_{{}^\circ }}=\sqrt{gH}\]
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