Q.

A particle of mass 2 kg moves with an initial velocity of v¯=4i^+4j^ms1. A constant force of F=20j^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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a

1.2 m

b

4.8 m

c

3.2 m

d

6.0 m

answer is D.

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Detailed Solution

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a=Fm=10j^ms12      Displacement in y direction

y=ut+12at20=4×t+12×10×t2

t=45sx=4t=4×45=3.2m

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