question_answer

A body A is thrown up vertically from the ground with a velocity \[{{\upsilon }_{0}}\]and another body B is simultaneously dropped from a height \[\frac{H}{2},\] They meet at a height \[\frac{H}{2},\]if \[{{\upsilon }_{0}}\]is equal to

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 \[\frac{H}{2}=\frac{1}{2}g\times \frac{v_{0}^{2}}{{{g}^{2}}}\]\[\Rightarrow \]\[H=\frac{v_{0}^{2}}{g}\] \[\Rightarrow \]\[{{v}_{0}}=\sqrt{gH}\]