A) \[t={{t}_{1}}-{{t}_{2}}\]
B) \[t=\frac{{{t}_{1}}+{{t}_{2}}}{2}\]
C) \[t=\sqrt{{{t}_{1}}{{t}_{2}}}\]
D) \[t=\sqrt{t_{1}^{2}-t_{2}^{2}}\]
Correct Answer: C
Solution :
From equation of motion, we have \[s=ut+\frac{1}{2}g{{t}^{2}}\] where, \[u\] is initial velocity, \[g\] the acceleration due to gravity and \[t\] the rime. For upward motion \[h=-u{{t}_{1}}-\frac{1}{2}gt_{1}^{2}\] ... (i) For downward motion \[h=-u{{t}_{2}}+\frac{1}{2}gt_{2}^{2}\] ... (ii) Multiplying Eq. (i) by \[{{t}_{2}}\] and Eq. (ii) by \[{{t}_{1}}\] and subtracting Eq. (ii) by Eq. (i), we get \[h({{t}_{2}}-{{t}_{1}})=\frac{1}{2}g{{t}_{1}}{{t}_{2}}({{t}_{2}}-{{t}_{1}})\] \[h=\frac{1}{2}g{{t}_{1}}{{t}_{2}}\] ... (iii) When stone is dropped from rest\[u=0\], reaches the ground in t second. \[\therefore \] \[h=\frac{1}{2}g{{t}^{2}}\] ... (iv) Equating Eqs. (iii) and (iv), we get \[\frac{1}{2}g{{t}^{2}}=\frac{1}{2}g{{t}_{1}}{{t}_{2}}\]
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