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A particle of mass 2kgmoves with an initial velocity of $(4 \hat{i} + 2 \hat{j}) {ms}^{- 1}$ on the x-y plane. A force $\vec{F} = (2 \hat{i} - 8 \hat{j}) N$ acts on the particle. The initial position of the particle is (2 m,3 m). Then for y=3 m,

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Possible value of x is only x = 2 m

Possible value of x is not only x=2 m, but there exists some other value of x also

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detailed solution

Correct option is B

u→=4i^+2j^, a→=F→m=i^−4j^Let at any time, the coordinates be (x, y)x−2=uxt+12axt2⇒x−2=4t+12t2 and y−3=2t−124t2⇒y−3=2t−2t2 When y=3m,t=0,1s; when t=0,x=2m When t=1s,x=6.5m

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A particle of mass 2kgmoves with an initial velocity of (4i^+2j^)ms−1 on the x-y plane. A force F→=(2i^−8j^)N acts on the particle. The initial position of the particle is (2 m,3 m). Then for y=3 m,

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A particle of mass 2kgmoves with an initial velocity of $(4 \hat{i} + 2 \hat{j}) {ms}^{- 1}$ on the x-y plane. A force $\vec{F} = (2 \hat{i} - 8 \hat{j}) N$ acts on the particle. The initial position of the particle is (2 m,3 m). Then for y=3 m,