A) \[\sqrt{2gH}\]
B) \[\sqrt{gH}\]
C) \[\frac{1}{2}\sqrt{gH}\]
D) \[\sqrt{\frac{2g}{H}}\]
Correct Answer: B
Solution :
Suppose the two bodies A and B meat at time At a height \[\frac{H}{2}\]from the ground. For body \[B,u=0,h=\frac{H}{2}\] \[\Rightarrow \]\[h=ut+\frac{1}{2}g{{t}^{2}}\]Hence,\[\frac{H}{2}=\frac{1}{2}g{{t}^{2}}\] ?(i) For body \[A,u={{v}_{0}},h\frac{H}{2}\] \[\Rightarrow \]\[h=ut-\frac{1}{2}g{{t}^{2}}\]\[\Rightarrow \]\[\frac{H}{2}={{v}_{0}}t-\frac{1}{2}g{{t}^{2}}\] (ii) So, from Eqs. (i) and (ii),\[{{v}_{0}}t-\frac{1}{2}g{{t}^{2}}=\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \]\[{{v}_{0}}t=g{{t}^{2}}\]\[\Rightarrow \]\[t=\frac{{{v}_{0}}}{g}\] Thus, we get \[\frac{H}{2}=\frac{1}{2}g\times \frac{v_{0}^{2}}{{{g}^{2}}}\]\[\Rightarrow \]\[H=\frac{v_{0}^{2}}{g}\] \[\Rightarrow \]\[{{v}_{0}}=\sqrt{gH}\]You need to login to perform this action.
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