Let the initial speed be $u$.

Since the body is projected at $45^\circ$, its initial velocity components are $u_x = u\cos45^\circ = \frac{u}{\sqrt{2}}, \qquad u_y = u\sin45^\circ = \frac{u}{\sqrt{2}}.$

After $t=2\,$s:

• Horizontal velocity remains unchanged: $v_x = \frac{u}{\sqrt{2}}$
• Vertical velocity becomes: $v_y = \frac{u}{\sqrt{2}} - gt = \frac{u}{\sqrt{2}} - 20$

The speed after 2 s is given as $20\,$m/s, so $v^2 = v_x^2 + v_y^2 = 20^2.$ Therefore, $\left(\frac{u}{\sqrt{2}}\right)^2 + \left(\frac{u}{\sqrt{2}} - 20\right)^2 = 400.$

Let $a = \frac{u}{\sqrt{2}}.$ Then, $a^2 + (a-20)^2 = 400.$ Expanding, $a^2 + a^2 - 40a + 400 = 400$ $2a^2 - 40a = 0$ $2a(a-20)=0.$

So, $a=0 \quad \text{or} \quad a=20.$ Since the projectile is actually launched, $a\neq 0$. Hence, $\frac{u}{\sqrt{2}}=20.$

Thus the initial vertical component is u_y = 20\,$m/s.

Maximum height is $H = \frac{u_y^2}{2g} = \frac{20^2}{2\cdot 10} = \frac{400}{20} = 20\,\text{m}.$

Therefore, the maximum height reached is $\boxed{20\,\text{m}}.$