A) \[\sqrt{2gH}\]
B) \[\sqrt{gH}\]
C) \[\frac{1}{2}\sqrt{gH}\]
D) \[\frac{2g}{H}\]
Correct Answer: B
Solution :
Suppose the two bodies A and B meet at time t, at a height \[\frac{H}{2}\] from the ground. For body \[B,\,u=0,\,\,h=\frac{H}{2}\] \[h=ut+\frac{1}{2}g{{t}^{2}}\] \[\frac{H}{2}=\frac{1}{2}g{{t}^{2}}\] ?. (i) For body A, \[u={{v}_{0}},h=\frac{H}{2}\] \[h=ut-\frac{1}{2}g{{t}^{2}}\] \[\frac{H}{2}={{v}_{0}}t-\frac{1}{2}g{{t}^{2}}\] .... (ii) From Eqs. (i) and (ii) \[{{v}_{0}}t-\frac{1}{2}g{{t}^{2}}=\frac{1}{2}g{{t}^{2}}\] or \[{{v}_{0}}t=g{{t}^{2}}\] or \[t=\frac{{{v}_{0}}}{g}\] Substituting the value of t in Eq. (i), we get \[\frac{H}{2}=\frac{1}{2}g{{t}^{2}}\] \[H=g\times {{\left( \frac{{{v}_{0}}}{g} \right)}^{2}}\] \[{{v}_{0}}=\sqrt{gH}\]You need to login to perform this action.
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