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

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$y = 2x - 5x^2$

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(B) The initial velocity vector is given by $\vec{u} = 1\hat{i} + 2\hat{j} \, m/s$.Comparing this with $\vec{u} = u_x\hat{i} + u_y\hat{j}$,we get $u_x = 1 \, m/s$ and $u_y = 2 \, m/s$.The horizontal component is $u_x = u \cos 	heta = 1 \, m/s$.The vertical component is $u_y = u \sin 	heta = 2 \, m/s$.The equation of the trajectory for a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Since $	an 	heta = \frac{u_y}{u_x} = \frac{2}{1} = 2$ and $u \cos 	heta = u_x = 1$,we substitute these 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 $(\hat{i} + 2\hat{j}) \, m/s$,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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