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(B) The Cartesian equation of the path of a projectile is given by $y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.Given the initial velocity

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$y=2 x-5 x^2$

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(B) The Cartesian equation of the path of a projectile is given by $y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.Given the initial velocity vector $\vec{u} = \hat{i} + 2 \hat{j} \,ms^{-1}$.Here, the horizontal component $u_x = 1 \,ms^{-1}$ and the vertical component $u_y = 2 \,ms^{-1}$.The horizontal displacement is $x = u_x t = 1 \cdot t$, so $t = x$.The vertical displacement is $y = u_y t - \frac{1}{2} g t^2$.Substituting $t = x$ into the equation for $y$:$y = 2(x) - \frac{1}{2} (10) (x)^2$.$y = 2x - 5x^2$.

$A$ projectile is given an initial velocity of $\hat{i}+2 \hat{j} \,ms^{-1}$. The Cartesian equation of its path is ($x$ and $y$ are in metres and $g=10 \,ms^{-2}$)

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