There is a uniform electric field of strength  ${10}^3 \mathrm{V}/\mathrm{m}$ along y-axis. A body of mass 1 g and charge ${10}^{-6} \mathrm{C}$ is projected into the field from origin along the positive x-axis with a velocity 10 m/s. Its speed in m/s after 10 s is (Neglect gravitation)

Body moves along the parabolic path.

For vertical motion : By using v = u + at

$v_y=0+\frac{QE}{m}\cdot t=\frac{{10}^{-0}\times {10}^3}{{10}^{-3}}\times 10=10\mathrm{m}/\sec$

For horizontal motion – It’s horizontal velocity remains the same i.e. after 10 sec, horizontal velocity of body v_x = 10 m/sec.

$v=\sqrt{v_x^2+v_y^2}=10\sqrt{2}\mathrm{m}/\sec$
Velocity after 10 sec