A particle moving in magnetic and gravitational fields.

NOTE: The particle travels in a helical motion. As it is not well known, I felt like proving it. So, here goes.

Let the particle begin its journey from $\left(0,H\right)$ (X-axis towards right, Y-axis upwards)

At a general position $\left(x,y\right)$ , let the Horizontal and Vertical Velocity of the particle be $v_x$ (directed towards right) and $v_y$ (directed downwards).

Also, let the Magnetic Field be directed towards $-\hat{k}$ (according to the adjusted coordinates).
The Gravitational force is directed downwards.

By applying Newton's laws separately for Horizontal and Vertical Direction, we have,

$\frac{\mathrm d v_x}{\mathrm d t} = \frac{qBv_y}{m}$ and $\frac{\mathrm d v_y}{\mathrm d t} = \frac{mg-qBv_x}{m}$

Now,
Differentiating the 2nd equation, and using the 1st equation,
$\frac{\mathrm d^2 v_y}{\mathrm d t^2} = \frac{-qB}{m}\left(\frac{\mathrm d v_x}{\mathrm d t}\right) = \frac{-qB}{m}\left(\frac{qB}{m} v_y\right)$

Clearly, this is an equation representing a Simple Harmonic Motion , and hence,
$v_y = A\sin(wt)$ , where $w=\frac{qB}{m}$

To find $A$ , we differentiate the above equation, to get,
$\frac{\mathrm d v_y}{\mathrm d t} = Aw\cos(wt)$

Substituting $t=0$ in the above equation, we get,
$\frac{\mathrm d v_y}{\mathrm d t} = Aw$

But, as initially, the vertical acceleration is $-g$ , we get,
$A = \frac{-g}{w}$

Using the value of $A$ , and Integrating the above equation to get $y$ , we finally have, $\int\limits_H^y dy = \int\limits_0^t \frac{-g}{w} \sin(wt) dt$

Finally, we have,
$y = H - \frac{g}{w^2}\left(1-\cos(wt)\right)$

Now, clearly, the Minimum value of $y$ occurs when $\cos(wt) = -1$
Hence,

$y_{min} = H - \frac{2g}{w^2}$

Setting $y_{min} = 0$ , substituting the value of $w$ , and using the values provided,
$H = 19.6$

Therefore, the minimum Integral Value of $H$ ,
$H_{min} = \boxed{20}$

NOTE: I found this so interesting, and hence, I want to notify that you can even calculate the time it takes for the ball to just reach the Earth. In fact, we have already done it ..if you notice carefully!

Such a beautiful question this is ,,, because though it is easy to solve,,, It is so beautiful,, It shows that we can actually have such a situation,, Thank you Jatin bro,,, that was such an innovative one, like all of your questions,, But this also means that the particles keep oscillating and in this limiting case ,, it just touches the ground and again goes up and back,,, What would make it harder and even more interesting is if g was not allowed to be constant :D