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. Possible value of x is only x=2m
B. Possible value of x is not only x=2m, but there exists some other value of x also
C. Time taken is 2s
D. All of the above

Answer

B

Explanation

`vec(u)=4hat(i)+2hat(j), vec(a)=(vecF)/(m)=hat(i) - 4hat(j)`
Let at any time, the coordinates be (x,y).
`x-2 =u_(x)t+1/2 a_(x)t^(2)`
`implies x-2 = 4t + 1/2 t^(2)` and `y-2 = 2t-1/2 4t^(2)`
`implies y-3 = 2t-2t^(2)`
when `y=3m, t=0, 1s`, when `t=0,x=2m`
when `t=1s, x=6.5m`.