Q91 / 180 · AIIMS 2017 · Physics

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

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

Solution

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

Hence, the correct answer is option: 'B'.

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