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 \[x=H-\frac{1}{2}g{{\left( \frac{H}{u} \right)}^{2}}\] \[x=H-\frac{1}{2}g\frac{{{H}^{2}}}{{{u}^{2}}}=H-\frac{1}{2}g\frac{{{H}^{2}}}{(3gH)}=H-\frac{H}{6}=\frac{5H}{6}\]You need to login to perform this action.
You will be redirected in
3 sec