A particle projected from a horizontal plane $(x - y$ plane $)$ such that its velocity vector at time $t$ is given by $\to V = a^i + (b - c t)^j$ . Its range on the horizontal plane is given by

\frac{b a}{c}

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\frac{2 b a}{c}

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\frac{3 b a}{c}

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N o n e

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Solution

The correct option is B $\frac{2 b a}{c}$
$\to v = a^i + (b - c t)^j$ So $, v_{x} = a v_{y} = b - c t$
$A t t = 0,$ initial velocity $u_{x} = a u_{y} = b$
Acceleration $a_{x} = 0 a_{y} = - c$
$y -$ co-ordinate of point $A$ is zero, so $0 = u_{y} t - \frac{1}{2} c t^{2} ⟹ 0 = b t - \frac{1}{2} c t^{2} ⟹ t = \frac{2 b}{c}$
Range along horizontal plane $= v_{x} t = \frac{2 b a}{c}$

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