Answer
$\vec v = (1.53\times10^{6}m/s) \hat{i} - (4.34 \times10^{5} m/s) \hat{j}$
Work Step by Step
$\vec{F} =q\vec{E} \Rightarrow m\vec{a}=q\vec{E}\Rightarrow ma=qE
\Rightarrow a =\frac{qE}{m}=\frac{ eE }{ m_{e}} = 8.78\times10^{11} m/s^{2}$
Because of uniform electric field, velocity will not change:
$\bullet v_{x}=v_{0x}=v_{0}\cos\theta=2 \times 10^{6} \cos 40^{\circ}=1.53 \times10^{6} m/s$
$\bullet v_{y}=v_{0y}=v_{0}\sin\theta- at$
We have: $ x=v_{x}t \Rightarrow t=\frac{x}{v_{x}}=\frac{3}{1.53 \times10^{6}}=1.96 \times 10^{-6}s $
Thus, $v_{y}=v_{0y}=v_{0}\sin\theta- at=(2 \times 10^{6} \sin 40^{\circ})-(8.78\times10^{11} m/s^{2})(1.96 \times 10^{-6})=-4.34 \times 10^{5} m/s$
Final Velocity is: $\vec v = (1.53\times10^{6}m/s) \hat{i} - (4.34 \times10^{5} m/s) \hat{j}$
