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

Options

A. 1.2m
B. 4.8m
C. 6.0m
D. 3.2m

Answer

D

Explanation

`vec(a) =(vec(F))/(m) = -10 hat(j) (ms^(-1))^(2)`
Displacement in y-direction
`y=ut + 1/2 at^(2) implies 0=4 xx t xx -1/2 xx 10 xx t^(2)`
`t=4/5 s implies x=4t = 4 xx 4/5 = 3.2m`