A horizontal cesium plate ( φ = 1.9 eV ) travels vertically...

The maximum energy of the incident photons corresponds to the minimum wavelength $\lambda = 250 \text{ nm}$.

$E = \dfrac{hc}{\lambda} = \dfrac{1240 \text{ eV nm}}{250 \text{ nm}} = 4.96 \text{ eV}$

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons relative to the plate is:

K_\max = E - \phi = 4.96 \text{ eV} - 1.9 \text{ eV} = 3.06 \text{ eV}

K_\max = 3.06 \times 1.6 \times 10^{-19} \text{ J} = 4.896 \times 10^{-19} \text{ J}

The maximum velocity of the photoelectrons relative to the plate is:

u_\max = \sqrt{\dfrac{2 K_\max}{m_e}} = \sqrt{\dfrac{2 \times 4.896 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 1.04 \times 10^6 \text{ m/s}

The velocity of a photoelectron relative to the ground is $\vec{v}_{e} = \vec{u} + \vec{v}_{plate}$, where $\vec{v}_{plate}$ is directed vertically downward with speed $v$.

For the vertically upward velocity component of every photoelectron to remain nonpositive, the downward speed of the plate $v$ must be greater than or equal to the maximum possible upward velocity of the photoelectron relative to the plate.

v \ge u_\max \Rightarrow v_\min = 1.04 \times 10^6 \text{ m/s}