Correct Answer - A::B
a,b. Since acceleration is in x-direction only, velocity in
y-direction will not change.
When speed `= 5 ms^(-1)`
`5^(2) = V_(x)^(2) + V_(y)^(2) = V_(x)^(2)+(4)^2`
`rArr V_x = +-3ms^(-1)`
`:. V_x = u_x + a_xt rArr t = (V_x-u_x)/(a_x)`
or `t_1 = (3-4)/(-0.4) = 2.5s`
and `t_2 = (3+4)/0.4 = 17.5s`.