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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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An object is thrown along a direction inclined at an angle of $45^{\circ}$ with the horizontal range of the particle is equal to

$\begin{array}{1 1}(A)\;vertical\; height\\(B)\;twice\;the \; vertical\;height \\(C)\;thrice \;the\; vertical \;height\\(D)\;four\; times \;the\;vertical\;height \end{array} $

Can you answer this question?
 
 

1 Answer

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$H= \large\frac{v_0^2 \sin ^245^{\circ}}{2g}$$R$
$R=4 H$
Hence D is the correct answer.
answered Jul 11, 2014 by meena.p
 

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