A) \[t={{t}_{1}}-{{t}_{2}}\]
B) \[t=\frac{{{t}_{1}}+{{t}_{2}}}{2}\]
C) \[t=\sqrt{{{t}_{1}}{{t}_{2}}}\]
D) \[t=t_{1}^{2}t_{2}^{2}\]
Correct Answer: C
Solution :
If a stone is dropped from height h then \[h=\frac{1}{2}g\,{{t}^{2}}\] ?(i) If a stone is thrown upward with velocity u then \[h=-u\ {{t}_{1}}+\frac{1}{2}g\ t_{1}^{2}\] ?(ii) If a stone is thrown downward with velocity u then \[h=u{{t}_{2}}+\frac{1}{2}gt_{2}^{2}\] ?(iii) From (i) (ii) and (iii) we get \[-u{{t}_{1}}+\frac{1}{2}g\,t_{1}^{2}=\frac{1}{2}g\,{{t}^{2}}\] ?(iv) \[u{{t}_{2}}+\frac{1}{2}g\,t_{2}^{2}=\frac{1}{2}g\,{{t}^{2}}\] ?(v) Dividing (iv) and (v) we get \[\therefore \]\[\frac{-u{{t}_{1}}}{u{{t}_{2}}}=\frac{\frac{1}{2}g({{t}^{2}}-t_{1}^{2})}{\frac{1}{2}g({{t}^{2}}-t_{2}^{2})}\] or \[-\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{t}^{2}}-t_{1}^{2}}{{{t}^{2}}-t_{2}^{2}}\] By solving \[t=\sqrt{{{t}_{1}}{{t}_{2}}}\]
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