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(B) The velocity vector is given by $\vec{V} = a\hat{i} + (b - ct)\hat{j}$.Comparing this with $\vec{V} = v_x\hat{i} + v_y\hat{j}$,we get $v_x = a$ an

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$\frac{2ba}{c}$

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(B) The velocity vector is given by $\vec{V} = a\hat{i} + (b - ct)\hat{j}$.Comparing this with $\vec{V} = v_x\hat{i} + v_y\hat{j}$,we get $v_x = a$ and $v_y = b - ct$.At $t = 0$,the initial velocity components are $u_x = a$ and $u_y = b$.The acceleration in the $y$-direction is $a_y = \frac{dv_y}{dt} = -c$.The time of flight $T$ is the time when the particle returns to the horizontal plane,i.e.,displacement in $y$ is $0$.Using $s_y = u_y T + \frac{1}{2} a_y T^2 = 0$,we get $bT - \frac{1}{2} cT^2 = 0$,which gives $T = \frac{2b}{c}$.The range $R$ is the displacement in the $x$-direction during the time of flight: $R = u_x T = a 	imes \left( \frac{2b}{c} ight) = \frac{2ab}{c}$.

$A$ particle is projected from a horizontal plane ($x-z$ plane) such that its velocity vector at time $t$ is given by $\vec{V} = a\hat{i} + (b - ct)\hat{j}$. Its range on the horizontal plane is given by:

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