Particle Leads Parabolic Track

Assume that the electric force is solely responsible for the trajectory. The position, velocity, and acceleration are:

$x = x \\ y = 3x - 4x^2 \\ \dot{x} = \dot{x} \\ \dot{y} = 3 \dot{x} - 8 x \dot{x} \\ \ddot{x } = \ddot{x} \\ \ddot{y} = 3 \ddot{x} - 8(x \ddot{x} + \dot{x}^2)$

At $x = 0$ :

$\ddot{x } = \ddot{x} \\ \ddot{y} = 3 \ddot{x} - 8 \dot{x}^2$

We know that the electric field is only in the $y$ direction, so this further reduces to:

$\ddot{x } = 0 \\ \ddot{y} = -8 \dot{x}^2$

Setting the force equal to the $y$ acceleration:

$m a = q E \\ 2021 (8 \dot{x}^2) = 2020 (2)$

This yields:

$|\dot{x}| \approx 0.5 \\ |\dot{y}| \approx 1.5 \\ \sqrt{\dot{x}^2 + \dot{y}^2} \approx 1.58$

Let the horizontal and the vertical components of the velocity of the particle at the origin be $u$ and $v$ respectively. There is no acceleration in the horizontal direction, and hence $u$ is constant. Acceleration of the particle along the downward vertical direction is $w=\dfrac{2020\times 2}{2021}$ m/s $^2$ . Hence the equation of the trajectory of the particle is $y=\frac{v}{u}x-\frac{1}{2}w\frac{x^2}{u^2}$ . Comparing with the given equation, we get $v=3u, u^2=\frac{2020\times 2}{8\times 2021}\implies$ the magnitude of velocity at the origin $=\sqrt {u^2+v^2}=u\sqrt {10}=\sqrt {\frac{5050}{2021}}\approx \boxed {1.58}$ . So the answer is $[1.58]=\boxed 2$ .