question_answer

A stone is dropped from the top of a tower of height H. At the same time a second stone is projected vertically with a velocity\[\sqrt{3gH}\]from the foot of the tower. At what height, from the foot of the tower, will the two meet?

A)  \[3H/4\]             

B)  \[5H/6\]

C)  \[3H/8\]            

D)  \[H/2\]

Correct Answer: B

Solution :

: For falling stone, \[(H-x)=(0\times t)+\frac{1}{2}g{{t}^{2}}\] Or \[H-x=\frac{1}{2}g{{t}^{2}}\]         ...(i) For ascending stone, \[x=(ut)-\frac{1}{2}g{{t}^{2}}\] ?.(ii)    \[\therefore \]From (i) and (ii) \[H=ut\] or \[t=\frac{H}{u}\]      ... (iii) \[\therefore \]From (i) \[H-\frac{1}{2}g{{t}^{2}}=x\] or \[x=H-\frac{1}{2}g{{\left( \frac{H}{u} \right)}^{2}}\] \[x=H-\frac{1}{2}g\frac{{{H}^{2}}}{{{u}^{2}}}=H-\frac{1}{2}g\frac{{{H}^{2}}}{(3gH)}=H-\frac{H}{6}=\frac{5H}{6}\]