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Question

A particle is uncharged and is thrown vertically upward from ground level with a speed of 25.6 m/s. As a result, it attains a maximum height h. The particle is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of 29.3 m/s. The electric potential at the hight h exceeds the electric potential at ground level. Finally, the particle is given a negative charge -1. Ignoring air resistance, determine the speed with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the maximum height h. In all three situations, be sure to include the effect of gravity.

Answer 1

I assume that this problem has not gotten a solution yet, because it cannot be solved as stated.

I am going to give you a solution on the assumption of a typo:

I assume that in the final experiment the charge is -q, not -1.

Let

m be the mass of the particle,

g be gravitational acceleration,

v1 = 25.6 m/s be the initial speed of the first experiment,

v2 = 29.3 m/s be the initial speed of the second experiment,

v3 (to be calculated) be the initial speed of the third experiment,

K1 = mv1²/2 be the initial kinetic energy of the first experiment,

K2 = mv2²/2 be the initial kinetic energy of the second experiment,

K3 = mv3²/2 be the initial kinetic energy of the third experiment,

P = mgh be the maximum gravitational potential energy reached in all three experiments,

E be the electrical potential energy of charge +q at height h with respect to the ground level.

Then the electrical potential energy of charge -q at height h with respect to the ground level is -E.

During the first experiment the initial kinetic energy K1

get transformed into the final gravitational potential energy P.

During the other two experiments the initial kinetic energy gets

transformed into the final gravitational potential energy and

the final electrical potential energy.

This is expressed by the equations

K1 = P

K2 = P + E

K3 = P - E

Add the last two equations, and substitute from the first to get

K2 + K3 = 2K1

Replace with definitions of kinetic energy

mv2²/2 + mv3²/2 = 2mv1²/2

Solve for

v3 = √(2v1² - v2²)

Substitute actual numbers

v3 = √(2*25.6² - 29.3²) = 21.3 m/s