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(B) The initial velocity is given by $\vec{u} = u_x \hat{i} + u_y \hat{j} = (1\hat{i} + 2\hat{j}) 	ext{ ms}^{-1}$.Thus,the horizontal component is $u

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$y = 2x - 5x^2$

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(B) The initial velocity is given by $\vec{u} = u_x \hat{i} + u_y \hat{j} = (1\hat{i} + 2\hat{j}) 	ext{ ms}^{-1}$.Thus,the horizontal component is $u_x = 1 	ext{ ms}^{-1}$ and the vertical component is $u_y = 2 	ext{ ms}^{-1}$.At any time $t$,the horizontal position is $x = u_x t = 1 \cdot t$,which implies $t = x$.The vertical position is $y = u_y t - \frac{1}{2}gt^2$.Substituting $g = 10 	ext{ ms}^{-2}$,$u_y = 2 	ext{ ms}^{-1}$,and $t = x$ into the equation:$y = 2x - \frac{1}{2}(10)x^2$.$y = 2x - 5x^2$.Therefore,the equation of the trajectory is $y = 2x - 5x^2$.

$A$ body is projected with a velocity $(\hat{i} + 2\hat{j}) \text{ ms}^{-1}$,where $\hat{i}$ is along the horizontal and $\hat{j}$ is vertically upward. Then the equation of its trajectory is $(g = 10 \text{ ms}^{-2})$.

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