A particle is projected from a horizontal plane such that its velocity vector at time t is given by

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3-2.Motion in Plane

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A particle is projected from a horizontal 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

A

$\frac{2ab}{c}$

B

$\frac{ab}{c}$

C

$\frac{3ab}{c}$

D

$\frac{4ab}{c}$

Solution

$v_{y}=0$ at $t=\frac{b}{c} \quad \therefore T=\frac{2 b}{c}$

Range $=\mathrm{u}_{\mathrm{x}} \mathrm{T}=\frac{2 \mathrm{ab}}{\mathrm{c}}$

Standard 11

Physics

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