Question

A projectile is given an initial velocity of $(\hat{i}+2\hat{j})\text{m/s}$. Then, the equation of its trajectory is: [Take $g=10{\text{m/s}}^2$]

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

Answer 1

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

We know equation of trajectory of a projectile is
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }...\left(1\right)$
Here, initial velocity is $\overrightarrow{u}=\hat{i}+2\hat{j}$


$\tan \theta =\frac{Y-\text{component}}{X-\text{component}}$
$\Rightarrow \tan \theta =\frac{2}{1}=2...\left(2\right)$
Magnitude of initial velocity,
$u=\sqrt{1^2+2^2}=\sqrt{5}\text{m/s}...\left(3\right)$
$\cos \theta =\frac{\text{base}}{\text{hypotenuse}}=\frac{1}{\sqrt{5}}$


On putting values in Eq. $\left(1\right)$,
$y=x\times 2-\frac{g{x}^2}{2\times (\sqrt{5}{)}^2\times {\left(\frac{1}{\sqrt{5}}\right)}^2}$
$\Rightarrow y=2x-\frac{10x^2}{2\times 5\times \frac{1}{5}}$
$\therefore y=2x-5x^2$
Equation of its trajectory is $y=2x-5x^2$