A) \[\frac{1}{{{t}_{3}}}=\frac{1}{{{t}_{2}}}-\frac{1}{{{t}_{1}}}\]
B) \[t_{3}^{2}=t_{1}^{2}-t_{2}^{2}\]
C) \[{{t}_{3}}=\frac{{{t}_{1}}+{{t}_{2}}}{2}\]
D) \[{{t}_{3}}=\sqrt{{{t}_{1}}{{t}_{2}}}\]
Correct Answer: D
Solution :
When stone is thrown up \[h=u{{t}_{1}}+\frac{1}{2}g{{t}_{1}}^{2}\] ?(i) When thrown down \[h=u{{t}_{2}}+\frac{1}{2}g{{t}_{2}}^{2}\] ?(ii) When released \[h=\frac{1}{2}g{{t}_{3}}^{2}\] ?(iii) \[\begin{align} & \underline{\begin{align} & \,\,\,\,\,h{{t}_{2}}=-\,u{{t}_{1}}\,{{t}_{2}}+\frac{1}{2}gt_{1}^{2}{{t}_{2}} \\ & \,\,\,\,\,h{{t}_{1}}=+u{{t}_{1}}{{t}_{2}}+\frac{1}{2}gt_{2}^{2}{{t}_{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \end{align}}\,\,\,\,\,\, \\ & h({{t}_{2}}+{{t}_{1}})=+\frac{1}{2}g{{t}_{1}}{{t}_{2}}({{t}_{1}}+{{t}_{2}}) \\ \end{align}\] \[h=\frac{1}{2}g{{t}_{1}}{{t}_{2}}\] ?(iv) Comparing (iii) and (iv) \[{{t}_{3}}=\sqrt{{{t}_{1}}{{t}_{2}}}\]
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