Question

A projectile is given an initial velocity of $\hat{i}+2\hat{j}$. The cartesian equation of its path is (Take $g=10m{s}^{-2})$.

• A. $y=x-5x^2$
• B. $y=2x-5x^2$
• C. $y=2x-15x^2$
• D. $y=2x-25x^2$

Answer 1

Answer: (B)

$y=2x-5x^2$

Given, $\overrightarrow{u}=\hat{i}+2\hat{j}=u_x\hat{i}+u_y\hat{j}$
Then, $u_x=1=u\cos \theta$
and $u_y=2=u\sin \theta$
$\therefore \tan \theta =\frac{u\sin \theta }{u\cos \theta }=\frac{2}{1}=2$
The equation of trajectory of a projectile motion is
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }=x\tan \theta -\frac{g{x}^2}{2(u\cos \theta {)}^2}$
$\therefore y=x\times 2-\frac{10\times x^2}{2(1{)}^2}=2x-5x^2$.