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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 Vy= 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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