Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
0 votes

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

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App