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(B) The initial velocity vector is given by $\vec{u} = (1 \hat{i} + 2 \hat{j}) \ ms^{-1}$.Comparing this with $\vec{u} = u_x \hat{i} + u_y \hat{j}$,we

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$y = 2x - 5x^2$

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(B) The initial velocity vector is given by $\vec{u} = (1 \hat{i} + 2 \hat{j}) \ ms^{-1}$.Comparing this with $\vec{u} = u_x \hat{i} + u_y \hat{j}$,we get the horizontal component $u_x = 1 \ ms^{-1}$ and the vertical component $u_y = 2 \ ms^{-1}$.The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u_x^2}$.Since $	an 	heta = \frac{u_y}{u_x} = \frac{2}{1} = 2$,we substitute the values into the equation:$y = x(2) - \frac{10 \cdot x^2}{2(1)^2}$$y = 2x - \frac{10x^2}{2}$$y = 2x - 5x^2$.

$A$ projectile is given an initial velocity of $(\hat i + 2\hat j) \ ms^{-1}$,where $\hat i$ is along the ground and $\hat j$ is along the vertical. If $g = 10 \ m/s^2$,the equation of its trajectory is

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