Aproton is released from rest at the positive plate of a parallel-plate capacitor. it crosses the capacitor and reaches the negative plate with a speed of 55000m/s . what will be the final speed of an electron released from rest at the negative plate?

The correct answer to this question is this one:

You will need these relations

qV = KE = 1/2mv^2

You know the charge of a proton (the elementary charge in this case), you know its mass, and you know it's velocity. From there you can find its kinetic energy and from there you can find it's voltage. Once you know the potential difference there all you have to do is reverse the process. Multiply the charge of an electron by V and then solve for the velocity using the mass of the electron

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.