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.