question_answer

A body is projected vertically upwards at time \[t=0\] and it is seen at a height \[H\]at timer, and \[{{t}_{1}}\] and \[{{t}_{2}}\] second during its flight. The maximum height attained is (g is acceleration due to gravity)

A)  \[\frac{g{{\left( {{t}_{2}}-{{t}_{1}} \right)}^{2}}}{8}\]      

B)  \[\frac{g{{\left( {{t}_{1}}-{{t}_{2}} \right)}^{2}}}{4}\]

C)  \[\frac{g{{\left( {{t}_{1}}-{{t}_{2}} \right)}^{2}}}{8}\]

D)  \[\frac{g{{\left( {{t}_{2}}-{{t}_{1}} \right)}^{2}}}{4}\]  

Correct Answer: C

Solution :

Let time taken by the body to fall from point C to B is t. Then \[{{t}_{1}}+2t={{t}_{2}}\] \[t=\left( \frac{{{t}_{2}}-{{t}_{1}}}{2} \right)\]         ?.(i) Total time taken to reach point C \[T={{t}_{1}}+t\] \[={{t}_{1}}+\frac{{{t}_{2}}-{{t}_{1}}}{2}\] \[=\frac{2{{t}_{1}}+{{t}_{2}}-{{t}_{1}}}{2}=\left( \frac{{{t}_{1}}+{{t}_{2}}}{2} \right)\] Maximum height attained \[{{H}_{\max }}=\frac{1}{2}g{{(T)}^{2}}\] \[=\frac{1}{2}g{{\left( \frac{{{t}_{1}}+{{t}_{2}}}{2} \right)}^{2}}\] \[=\frac{1}{2}g.\frac{{{({{t}_{1}}+{{t}_{2}})}^{2}}}{4}\] \[\Rightarrow \] \[{{H}_{\max }}=\frac{1}{8}g.{{({{t}_{1}}+{{t}_{2}})}^{2}}m\]