question_answer

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

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                 \[\frac{H}{2}=\frac{1}{2}g{{t}^{2}}\]                 \[H=g\times {{\left( \frac{{{v}_{0}}}{g} \right)}^{2}}\]                 \[{{v}_{0}}=\sqrt{gH}\]