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A rocket is launched with velocity 10 km/s if radius of earth is R, then maximum height attained by it will be

$(a)\;2\; R \quad (b)\;4\;R \quad (c)\; 3\;R \quad (d)\;5\;R$

$KE$ given =$P.E$ of the body
$\large\frac{1}{2} $$mv^2=mg (R+h) v^2=2g(R+h) Where 'h' maximum height reached by the satellite \bigg( 10 \times \large\frac{5}{8}\bigg)^2$$\times 10^6= 2\times 10(R+h)$
Solving we get
$R+h=5R$
Where $R= 6.4 \times 10^6 \;m$
$R+h=5R$
$\quad h=4R$
Hence b is the correct answer.

edited Feb 17, 2014 by meena.p