A particle leaves the origin with a speed of 28 times 106 ms

A particle leaves the origin with a speed of 2.8 times 106 m/s at 28 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x = 1.5 cm if the particle is an electron.

Solution

u = 2.8*10^6 m/sec

angle = 28 deg

distance x = 1.5 cm = 1.5*10^-2 m

from the projectile motion the horizontal distance travelled by electron is

x = u*cos A*t

t = x/(u*cos A)

F = qE

ma = qE

a = qE/m

y = u*sinA*t - 0.5*a*t^2

y = u*sinA*t - 0.5*(qE/m)*t^2

if y = 0 then

t = 2mu*sinA/(qE) = x/(u*cosA)

E = 2mu^2*sinA*cosA/(x*q)

E = 2*9.1*10^-31*2.8^2*10^12*(sin 28 deg)*(cos 28 deg)/(1.5*10^-2*(-1.6)*10^-19)

E = - 2464.45 N/C

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