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(C) Given: Initial velocity $\vec{u} = 5 \hat{j} \, m/s$,acceleration $\vec{a} = 10 \hat{i} + 4 \hat{j} \, m/s^2$,and initial position $(x_0, y_0) = (

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$2 \, s$ and $18 \, m$

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(C) Given: Initial velocity $\vec{u} = 5 \hat{j} \, m/s$,acceleration $\vec{a} = 10 \hat{i} + 4 \hat{j} \, m/s^2$,and initial position $(x_0, y_0) = (0, 0)$.The position vector at time $t$ is given by $\vec{r} = \vec{u}t + \frac{1}{2} \vec{a}t^2$.For the $x$-coordinate:$x = u_x t + \frac{1}{2} a_x t^2$$20 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2$$20 = 5t^2 \implies t^2 = 4 \implies t = 2 \, s$.For the $y$-coordinate:$y = u_y t + \frac{1}{2} a_y t^2$$y_0 = 5 \cdot t + \frac{1}{2} \cdot 4 \cdot t^2$Substituting $t = 2 \, s$:$y_0 = 5(2) + 2(2^2) = 10 + 8 = 18 \, m$.Thus,$t = 2 \, s$ and $y_0 = 18 \, m$.

Starting from the origin at time $t=0,$ with initial velocity $5 \hat{j} \, m/s,$ a particle moves in the $x-y$ plane with a constant acceleration of $(10 \hat{i} + 4 \hat{j}) \, m/s^2$. At time $t$,its coordinates are $(20 \, m, y_0 \, m)$. The values of $t$ and $y_0$ are respectively:

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