two small ball having the same mass and charge and located at on same vertical at height div and H2 shown in the same direction along a horizontal at same velocity v and first ball thrown the ground at a distance l from initial what is his at what height will the second wall be at the instant the air dragand the charges and induced should be neglected

let mass of both the charges is m and charge on both the charges is q

take a sphere which is at height h1 two forces act on it

force of gravitation in downword direction and electrostatic force in downward direction hence net downward direction is qe+mg/m hence accelaration is qe/m+g hence

h1=1/2(g+qe/m)t^2

let t is time taken by small sphere in travelling distance

l=vt , t=l/v

let x is distance travelled by second sphere

forces acting on it

force of gravitation downward and electrostatic force upwards hence net acceleration in downward direction is( g-qe/m)

X=1/2(g-qe/m)t^2

height at this instant is h2-x

x=gl^2/v^2-h1

hence height is h2-(gl^2/v^2)

Regards

Arun (askIITians forum expert)