Two small balls having the same mass and charge and located on the same vertical at heights `h_(1)` and `h_(2)` are thrown in the same direction along the horizontal at the same velocity v. The first vall touches the ground at a horizontal distance R from the initial vertical position. At waht heigth `h_(2)` will the second ball be at this instant? Neglect any frictional resistance of air and the effect of any induced charge on the ground.

Answer

`h_(1)+h_(2)-g((R)/(v))^(2)` .

Explanation

Our system includes the two balls. The Coulomb force between the two balls is an internal force for the system. Internal force do not affect the motion of the center of mass. The motion of center of mass moves along a parabolic trajectory. Since the initial velocity of two balls is horizontal the time taken to travel distace x is `x//v` and the vertical height fallen by the center of mass in this time is
`y=(1)/(2)g((x)/(v))^(2)`
Position of center of mass at this moment from the ground is

`h=(h_(1)+h_(2))/(2)-(1)/(2)g((x)/(v))^(2)`
When the first ball touches the ground at a distance x=R, the height of the center of mass from the ground is
`h=(h_(1)+h_(2))/(2)-(1)/(2)g((R)/(v))^(2)`
As the masses of the balls are equal, the second ball will be at a height `h_(2)=2h` at this instant. Therefore,
`h_(2)=(h_(1)+h_(2))-g((R)/(v))^(2)` .