Question

A proton gun fires a proton from midway between two plates, A and B, which are separated by a distance of 10.7 cm; the proton initially moves at a speed of 145 km/s toward plate B. Plate A is kept at zero potential, and plate B at a potential of 759 V. With what speed will it hit plate A?

Answer 1

Well you can calculate it with conservation of energy or simple kinematics.
WITH ENERGY
Now you should know, that between two plates there is a homogeneous conservative electric field.
Which means, that potential at any point is proportional to its distance form the 0 potential plate.
V=E*d, where
E - electric field intensity, d- distance from plate 0 plate.
E=V/d=759/0.1=7590 V/m
Therefore proton at the beginning has ~406V potential.
First of all you find out what energy the proton has
m=1.7*10^-27 kg
q=1.6*10^-19 C
v=15*10^4 m/s
E1=KE+PE=mv^2/2 + qU=1.7*10^-27*(15*10^4)^2/2 +1.6*10^-19*406= 3.825*^-17 +  6.496*^-17 = 1.0321*^-16 J

When hitting plate A, the proton will have 0 potential energy, therefore everything is stored in speed.
E2=m*v^2/2
E2=E1
v=sqrt(2*E1/m)=sqrt(2*1.0321*^-16/(1.7*10^-27) ~ 1.101 km/s

Calculations maybe wrong, logic is correct!!

Hope this helps :)