Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Motion in a Plane

A projectile is given an initial velocity of $\hat i+2 \hat j$. The cartesian equation of its path is $(g=10 m/s^2)$. ($\hat i$ vector along horizontal $\hat j$ vector along vertically upward)

\[(a)\;y=2x-5x^2 \quad (b)\;y=x-5x^2 \quad (c)\;4y=2x-5x^2\quad (d)\;y=2x-25 x^2\]

1 Answer

From given velocity
$\tan {\theta}=\large\frac{y \; componet}{x \;component}$
Therefore $ \cos \theta=\large\frac{1}{\sqrt 5}$
$y=x \tan \theta-\large\frac{gx^2}{2u^2\cos ^2 \theta}$
$\quad= x(2)-\large\frac{10 (x^2)}{2\bigg( \sqrt {2^2+1^2}\bigg)^2 \bigg(\Large\frac{1}{\sqrt 5}\bigg)^2}$
Hence a is the correct answer.


answered Jul 1, 2013 by meena.p
edited Jan 26, 2014 by meena.p

Related questions