A) \[\frac{g{{({{t}_{2}}-{{t}_{1}})}^{2}}}{8}\]
B) \[\frac{g{{({{t}_{2}}+{{t}_{1}})}^{2}}}{4}\]
C) \[\frac{g{{({{t}_{2}}+{{t}_{1}})}^{2}}}{8}\]
D) \[\frac{g{{({{t}_{2}}-{{t}_{1}})}^{2}}}{4}\]
Correct Answer: C
Solution :
Let the speed of the body be v when it is at a height H above the ground. \[v=v-g{{t}_{1}}\] \[-v=u-g{{t}_{2}}\] Adding the two relations, we get \[0=2u-g({{t}_{1}}+{{t}_{2}})\Rightarrow u=\frac{g({{t}_{1}}+{{t}_{2}})}{2}\] Maximum height \[=\frac{{{g}^{2}}}{2g}=\frac{g{{({{t}_{1}}+{{t}_{2}})}^{2}}}{8g}\] \[=\frac{g{{({{t}_{1}}+{{t}_{2}})}^{2}}}{8}\] Hence, the correction option is [c].
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