Question

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

• A. $y=2x-5x^2$
• B. $y=x-5x^2$
• C. $4y=2-5x^2$
• D. $y=2x-25x^2$

Answer 1

The correct option is A $y=2x-5x^2$

Initial velocity of projectile = $\hat{i}+2\hat{j}$
$\Rightarrow u_x=1,u_y=2$
We know that Range of a projectile
$R=\frac{u^2\sin 2\theta }{g}=\frac{2\left(u\cos \theta \right)\left(u\sin \theta \right)}{g}$
$\Rightarrow R=\frac{2u_x{u}_y}{g}=\frac{2\times 1\times 2}{10}=\frac{2}{5}\text{m}$
$\tan \theta =\frac{u_y}{u_x}=\frac{2}{1}=2$
Using equation of trajectory of a projectile, we get
$y=x\tan \theta (1-\frac{x}{R})$
$y=2x(1-\frac{5x}{2})=2x-5x^2$
Alternative solution:
$u_x=1,u_y=2$
$\tan \theta =\frac{u_y}{u_x}=2$
And $u=\sqrt{5}\text{m/s}$
From equation of trajectory
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$
$y=2x-\frac{10x^2}{2\times 5\times {\left(\frac{1}{\sqrt{5}}\right)}^2}$
$y=2x-5x^2$