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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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A particle is projected from the surface of earth with an initial speed of 4 km/s . Find the maximum height attained by the particle.Radius of earth =6400 km and $g= 9.8 m/s^2$

\[(a)\;1500 \;km \quad (b)\; 858\; km \quad (c)\;355\;km \quad (d)\;935\;km \]
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Decrease in Kinetic energy = increase in gravitational potential energy of particle
$\large\frac{1}{2}$$ mv^2 =\large\frac{mgh}{1+\Large\frac{h}{R}}$
$\therefore h= \large\frac{v^2}{2g-\Large\frac{v^2}{R}}$
$\qquad=\large\frac{(4.0 \times 10 ^3)^2}{2 \times 9.8 -\Large\frac{(4.0 \times 10 ^3)^2}{6.4 \times 10^6}}$
$\qquad=9.35 \times 10 ^5 \;m$
we get $ h= 935 \;km$
Hence d is the correct answer.
answered Aug 20, 2013 by meena.p
edited Jun 29, 2014 by lmohan717

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