In a uniform horizontal magnetic field with induction B = 10mTl falls freely perpendicular to the field induction

In a uniform horizontal magnetic field with induction B = 10mTl falls freely perpendicular to the field induction line B (vector), a horizontally rectilinear conductor with a length of l = 0.8 m is placed.Find after what time interval (delta t) from the beginning of the fall of the conductor EMF of induction at its ends will be 49 mV

When a conductor moves perpendicular to the lines of induction of a magnetic field, the EMF that occurs in it is determined by the equation:
E = B * l * v,
Where
E = 49mV = 0.049V – EMF,
B = 10mT = 0.01T – magnetic field induction,
l = 0.8m – conductor length,
and v is the speed of movement of the conductor.
v = E / (B * l);
Free fall
v = g * (t ^ 2) / 2,
where g = 9.8m / s ^ 2 is the acceleration of gravity.
You need to find t:
t ^ 2 = 2 * v / g = 2 * E / (B * l * g) = 2 * 0.049 / (0.01 * 0.8 * 9.8) = 1.25;
t = 1.118s