question_answer

A stone is dropped from a height \[h\] simultaneously, another stone is thrown up from the ground which reaches a height\[4h\]. The two stones cross each other after time

A) \[\sqrt{\frac{h}{8g}}\]                   

B) \[\sqrt{8gh}\]

C) \[\sqrt{2gh}\]                   

D) \[\sqrt{\left( \frac{h}{2g} \right)}\]

Correct Answer: A

Solution :

For first stone\[v=0\]and For second stone\[\frac{{{v}^{2}}}{2g}=4h\]\[\Rightarrow \]\[{{u}^{2}}=8gh\] \[\therefore \]  \[u=\sqrt{8gh}\] Now,     \[{{h}_{1}}=\frac{1}{2}g{{t}^{2}}\]                 \[{{h}_{2}}=\sqrt{8ght}-\frac{1}{2}g{{t}^{2}}\] where, t=time to cross each other. \[\therefore \]  \[{{h}_{1}}+{{h}_{2}}=h\] \[\Rightarrow \]\[\frac{1}{2}g{{t}^{2}}+\sqrt{8ght}-\frac{1}{2}g{{t}^{2}}=h\] \[\Rightarrow \]               \[t=\frac{h}{\sqrt{8gh}}=\sqrt{\left( \frac{h}{8g} \right)}\]