A particle is uncharged and is thrown vertically upward from ground level with a

Question

A particle is uncharged and is thrown vertically upward from ground level with a speed of 26.3 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 32.5 m/s. The electric potential at the height h exceeds the electric potential at ground level. Finally, the particle is given a negative charge q. 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.

Explanation / Answer

Particle of mass m and velocity v = 26.3 m/s
KE = ½mv(1)²
PE = mgh
h = v(1)²/2g = (26.3)²/2(9.8) = 35.3 m

Now the particle has charge q+, and v(2) = 32.5 m/s
KE = ½mv(2)²
PE = mgh + kqQ/h (Q is the electrical charge of the Earth)
½mv(2)² = mgh + kqQ/h
½mv(2)² - mgh = kqQ/h
m = kqQ/h(½v(2)² - gh)

If the particle has -q, then
KE = ½mv²
PE = mgh - kqQ/h
½mv² = mgh - kqQ/h
v = sqrt[ 2gh - 2kqQ/mh]

Now substitue the expression we have for m into the above equation

v = sqrt[ 2gh - 2kqQ/(kqQ/h(½v(2)² - gh))h]
= sqrt[ 2gh - v(2)² + 2gh]
= sqrt[ 4gh - v(2)² ]
= sqrt[1383.76 - 1056.25 ]
= 18.1 m/s

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