question_answer

A body thrown vertically up to reach its maximum height in t second, the total time from the time of projection to reach a point at half of its maximum height while returning (in sec) is:

A)  \[\sqrt{2}\,t\]                            

B)  \[\left( 1+\frac{1}{\sqrt{2}} \right)t\]  

C)  \[\frac{3\,t}{2}\]                                  

D)  \[\frac{t}{\sqrt{2}}\]

Correct Answer: B

Solution :

                        \[t'\,\sqrt{\frac{2\left( \frac{h}{2} \right)}{g}}=\,\,\sqrt{\frac{h}{g}}=\,\sqrt{\left( \frac{\frac{1}{2}g{{t}^{2}}}{g} \right)}\,=\frac{t}{\sqrt{2}}\] Required time \[=t+t'\,=t+\frac{t}{\sqrt{2}}=t\left( 1+\frac{1}{\sqrt{2}} \right)\]