Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:

Explanation:
From the question we are told that:
Electric field 
Distance 
At negative plate
Generally the equation for Velocity is mathematically given by

Therefore




Answer:
H = start height (v = 0)
h = present height
v = present speed
assuming no friction
total energy = PE + KE
mgH = mgh + .5mv^2
if PE = KE then
mgH = mgh + mgh
h = H/2
potential energy = kinetic energy when object is at half its start height.
Explanation:
Answer:
27 m/s
Explanation:
Given:
v₀ = 15 m/s
a = 3 m/s²
t = 4 s
Find: v
v = at + v₀
v = (3 m/s²) (4 s) + (15 m/s)
v = 27 m/s