What must be the distance in meters between point charge q1 = 23.5 µc and point charge q2 = -64.2 µc for the electrostatic force between them to have a magnitude of 4.67 n?

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft