Aproton moves through a region of space where there is a magnetic field b = (0.45i + 0.38j) t and an electric field e = (3.0i - 4.2j) x 10^3 v/m. at a given instant, the proton's velocity is v = (6.0i + 3.0j - 5.0k) x 10^3m/s. determine the components of the total force on the proton.

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