A) \[t={{t}_{1}}-{{t}_{2}}\]
B) \[t=\frac{{{t}_{1}}+{{t}_{2}}}{2}\]
C) \[t=\sqrt{{{t}_{1}}{{t}_{2}}}\]
D) \[t_{1}^{2}=t_{2}^{2}\]
Correct Answer: C
Solution :
| [c] For first case of dropping \[h=\frac{1}{2}g{{t}^{2}}\] |
| For second case of downward throwing, |
| \[h=-u{{t}_{1}}+\frac{1}{2}gt_{1}^{2}=\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \] \[-u{{t}_{1}}=\frac{1}{2}g({{t}^{2}}-t_{1}^{2})\] |
| For third case of upward throwing, |
| \[h=u{{t}_{2}}+\frac{1}{2}gt_{2}^{2}=\frac{1}{2}g{{t}^{2}}\]\[\Rightarrow \] \[u{{t}_{2}}=\frac{1}{2}g({{t}^{2}}-t_{2}^{2})\] |
| On solving these two equations: \[-\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{t}^{2}}-t_{1}^{2}}{{{t}^{2}}-t_{2}^{2}}\] |
| \[\Rightarrow \] \[t=\sqrt{{{t}_{1}}{{t}_{2}}}\] |
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