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2.Motion in Straight Line
hard
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:
A
$gH=(n-2)^2u^2$
B
$2gH=nu^2(n-2)$
C
$gH=(n-2)u^2$
D
$2gH=n^2u^2$
(JEE MAIN-2014)
Solution
$\begin{array}{l}
Speed\,on\,reaching\,ground\,v = \sqrt {{u^2} + 2gh} \\
Now,\,v = u + at\\
\Rightarrow \,\,\,\,\sqrt {{u^2} + 2gh} = – u + gt\\
Time\,taken\,to\,reach\,highest\,{\rm{point}}\,is\,t = \frac{u}{g},\\
\Rightarrow t = \frac{{u + \sqrt {{u^2} + 2gH} }}{g} = \frac{{nu}}{g}\left( {from\,question} \right)\\
\Rightarrow 2gH = n\left( {n – 2} \right){u^2}
\end{array}$
Standard 11
Physics
