question_answer

From a tower of height H, a particle is thrown vertically upwards with a speed U. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path.            [JEE MAIN 2014]

The relation between H, u and n is:

A) \[2gH=n{{u}^{2}}(n-2)\]

B) \[gH=(n-2){{u}^{2}}\]

C) \[2gH={{n}^{2}}{{u}^{2}}\]

D) \[gH={{(n-2)}^{2}}{{u}^{2}}\]

Correct Answer: A

Solution :

[a] Time taken to reach highest point

For time taken to reach the ground\[{{t}_{1}}=\frac{u}{g}\]

\[-H=ut-\frac{u}{g}g{{t}^{2}}\] \[\therefore \]\[g{{t}^{2}}-2ut-2H=0\]

\[t=\frac{2u\pm \sqrt{4{{u}^{2}}+8gH}}{2g}\]

\[t=\frac{+\sqrt{{{u}^{2}}+2gH}}{g}\]= (− ve sign not acceptable)

Given \[t=n\,{{t}_{1}}\]                    \[\frac{u+\sqrt{{{u}^{2}}+2gH}}{g}=\frac{nu}{g}\]

Solving \[2gH=n{{u}^{2}}(n-2)\]