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A particle of mass 2 kg moves with an initial velocity of $\bar{v} = 4 \hat{i} + 4 \hat{j} {ms}^{- 1}$ . A constant force of $\vec{F} = - 20 \hat{j} N$ is applied on the particle. Initially, the particle was at 0, 0 The x-coordinate of the particle when its y-coordinate again becomes zero is given by

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1.2 m

4.8 m

3.2 m

6.0 m

answer is D.

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Detailed Solution

$\vec{a} = \frac{\vec{F}}{m} = - 10 \hat{j} {({ms}^{- 1})}^{2}$ Displacement in y direction

$y = ut + \frac{1}{2} {at}^{2} \Rightarrow 0 = 4 \times t + \frac{1}{2} \times 10 \times t^{2}$

$t = \frac{4}{5} s \Rightarrow x = 4 t = 4 \times \frac{4}{5} = 3 .2 m$

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