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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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If initial velocity of projection is $10 m/s$ and angle of projection is $60 ^{\circ}$, find range R.

\[(a)\;\frac{15 \sqrt 3}{2} \; m \quad (b)\;\frac{40}{3}m \quad (c)\;5 \sqrt 3\; m\quad(d)\;\frac{20}{3}\;m \]

Can you answer this question?
 
 

1 Answer

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For a projection on an inclined plane
Time of flight $T= \large\frac{2u \sin ( \alpha-\beta)}{y \cos \beta}$
$OA=Range=\large\frac{2 u^2 \sin ( \alpha- \beta) \cos \alpha}{g \cos ^2 \beta}$
Therefore $Range= \large\frac{2 \times 10 ^2 \sin (60-30) \cos 60}{10 \times \cos ^2 30}$
$\qquad= \large\frac{2 \times 10 ^2 \times \Large\frac{1}{2} \times \Large\frac{1}{2}}{10 \times \Large\frac{3}{4}}$
$\qquad= \large\frac{20}{3} \;$$m$
Hence d is the correct answer.

 

answered Jul 3, 2013 by meena.p
edited Jan 26, 2014 by meena.p
 

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