As the motion under gravity is symmetric, so distance travelled in last second of ascent is equal to first second of descent.

$t=1 s \left(1^{\text {st }} \text { second }\right)v=0$

$-x_2=u t-\frac{1}{2} g \times 1^2 x_1+x_2$

$x_2=\frac{1}{2} \times 9.8 \times 1^2(\because u=0)$

$\Rightarrow x_2=4.9 \,m$

This distance is constant for every body thrown with any speed.