JEE Mains · Physics · STD 11 - Motion in straight line
- JEE Mains 2014
- Hard
- MCQ
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. The relation between \(H, u\) and \(n\) is:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(2gH=nu^2(n-2)\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
Speed\,on\,reaching\,ground\,v = \sqrt {{u^2} + 2gh} \\
Now,\,v = u + at\\
\Rightarrow \,\,\,\,\sqrt {{u^2} + 2gh} = - u + gt\\
Time\,taken\,to\,reach\,highest\,{\rm{point}}\,is\,t = \frac{u}{g},\\
\Rightarrow t = \frac{{u + \sqrt {{u^2} + 2gH} }}{g} = \frac{{nu}}{g}\left( {from\,question} \right)\\
\Rightarrow 2gH = n\left( {n - 2} \right){u^2}
\end{array}\)

Speed\,on\,reaching\,ground\,v = \sqrt {{u^2} + 2gh} \\
Now,\,v = u + at\\
\Rightarrow \,\,\,\,\sqrt {{u^2} + 2gh} = - u + gt\\
Time\,taken\,to\,reach\,highest\,{\rm{point}}\,is\,t = \frac{u}{g},\\
\Rightarrow t = \frac{{u + \sqrt {{u^2} + 2gH} }}{g} = \frac{{nu}}{g}\left( {from\,question} \right)\\
\Rightarrow 2gH = n\left( {n - 2} \right){u^2}
\end{array}\)

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