A) \[\frac{g}{4}{{({{t}_{1}}+{{t}_{2}})}^{2}}\]
B) \[g{{\left( \frac{{{t}_{1}}+{{t}_{2}}}{4} \right)}^{2}}\]
C) \[2g{{\left( \frac{{{t}_{1}}+{{t}_{2}}}{4} \right)}^{2}}\]
D) \[\frac{g}{4}({{t}_{1}}{{t}_{2}})\]
Correct Answer: C
Solution :
Body crosses twice means it takes time\[{{t}_{1}}\]during upward journey and crosses the same point during downward journey. Time taken by the body to reach the point A is \[{{t}_{1}}\]and to come again to A it takes time \[({{t}_{2}}-{{t}_{1}})\]time taken to reach max. height is\[t={{t}_{1}}\left( \frac{{{t}_{2}}+{{t}_{1}}}{2} \right)\] [\[\because \]Time of ascent = Time of descent] \[t=\frac{{{t}_{2}}+{{t}_{1}}}{2}\] so max. height\[H=\frac{1}{2}g{{t}^{2}}\] \[=\frac{1}{2}g{{\left( \frac{{{t}_{1}}+{{t}_{2}}}{2} \right)}^{2}}\] \[=2g{{\left( \frac{{{t}_{1}}+{{t}_{2}}}{4} \right)}^{2}}\]
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