Dear tanya

since no force in the horizontal direction so horiizonta velocity will reamain constant Vx = 3.82x10⁷ m/s

so time taken to travel 0.066 m horixontally is  t = .066/3.82x10⁷ = 1.73  X 10^-9 s

charge of the proton is q = 1.6 * 10^-19C

mass  of the proton is m = 1.67 * 10^-27 kg

given E =20000 N/C

 acceleration of proton in vertical direction  a=qE/m

                                                                               =1.916 * 10¹²  m/sec²

vertical distance moved by proton  h= 1/2  at²

                                                                   =1/2  * 1.916 * 10¹² *(1.73  X 10^-9)²

                                                                   =2.867 * 10^-6 m

 vertical velocity of proton after time  t   V_y= at

                                                                            =1.916 * 10¹² *(1.73  X 10^-9)

                                                                             =3.315 * 10³ m/sec

so resultanat velocity will be  = √Vx² + Vy²

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