Vs - velocity on beginning
ve - velocity on ending. You've got:

So he needed 4 second.
Clouds and fog and it may rain or snow
It would be helpful if you draw the figure of the problem. You will see that a right triangle would be constructed by the problem where 19.0 is the angle between the hypotenuse and the base of the triangle. It is said that the force acting is said to be 9.0 N at the said angle to the horizontal. Using trigonometric relations,
cos 19 = adjacent / hypotenuse = horizontal component / 9
horizontal component = 8.51 N
Answer:
E₂ / E₁ = 521.64 / 5.95 =87.67
Explanation:
Let d be the distance covered inside electric field . Lt q be the magnitude of charge.
Force under field E₁ = q E₁
acceleration = qE₁/ m
d = 1/2 a t²
d = .5 ( qE₁ / m) x 32.3²
d = 521.64 ( qE₁ / m)
Similarly for return journey,
d = .5 x ( qE₂ / m) x 3.45²
d = 5.95x( qE₂ / m)
521.64 ( qE₁ / m) = 5.95x( qE₂ / m)
E₂ / E₁ = 521.64 / 5.95 =87.67