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A projectile is given an initial velocity of $\hat{i}+2 \hat{j} \mathrm{~ms}^{-1}$. The cartesian equation of its path is ( $x$ and $y$ are in metres and $\mathrm{g}=10 \mathrm{~ms}^{-1}$ )
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The correct answer is:
$y=2 x-5 x^2$
Cartesian equation path of projectile is given by
$y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}$
Given velocity $u=\hat{i}+2 \hat{j} \mathrm{~m} / \mathrm{s}$
$\begin{aligned} & |\mathrm{u}|=\sqrt{1^2+2^2}=\sqrt{5}, \tan \theta=\frac{2}{1}=2 \\ & \cos \theta=\frac{1}{\sqrt{5}} \\ & \therefore \mathrm{y}=\mathrm{x} \times 2-\frac{\mathrm{gx}^2}{2 \times(\sqrt{5})^2 \times\left(\frac{1}{\sqrt{5}}\right)^2}=2 \mathrm{x}-\frac{10 \times \mathrm{x}^2}{10 \times \frac{1}{5}} \\ & \therefore \mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^2 .\end{aligned}$
$y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}$
Given velocity $u=\hat{i}+2 \hat{j} \mathrm{~m} / \mathrm{s}$
$\begin{aligned} & |\mathrm{u}|=\sqrt{1^2+2^2}=\sqrt{5}, \tan \theta=\frac{2}{1}=2 \\ & \cos \theta=\frac{1}{\sqrt{5}} \\ & \therefore \mathrm{y}=\mathrm{x} \times 2-\frac{\mathrm{gx}^2}{2 \times(\sqrt{5})^2 \times\left(\frac{1}{\sqrt{5}}\right)^2}=2 \mathrm{x}-\frac{10 \times \mathrm{x}^2}{10 \times \frac{1}{5}} \\ & \therefore \mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^2 .\end{aligned}$
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