Up and down motion

We throw the body from a height h = 5 m above the Earth vertically upwards v0 = 10 m/s. How long before we have to let the second body fall freely from the same height to hit the Earth at the same time?

Correct result:

t =  1.4445 s

Solution:

$$\begin{gathered} h=5\text{ m } \\ {v}_0=10\text{ m/s } \\ g=9.81{\text{m/s}}^2 \\ {t}_1=v_0\mathrm{/}g=10\mathrm{/}9.81={\displaystyle \frac{1000}{981}}\doteq 1.0194\text{ s } \\ {h}_1=h+{\displaystyle \frac{1}{2}}\cdot g\cdot t_1^2=5+{\displaystyle \frac{1}{2}}\cdot 9.81\cdot 1.0194^2={\displaystyle \frac{9905}{981}}\doteq 10.0968\text{ m } \\ {h}_1={\displaystyle \frac{1}{2}}\cdot g\cdot t_2^2 \\ {t}_2=\sqrt{2\cdot h_1\mathrm{/}g}=\sqrt{2\cdot 10.0968\mathrm{/}9.81}\doteq 1.4347\text{ s } \\ {t}_3=\sqrt{2\cdot h\mathrm{/}g}=\sqrt{2\cdot 5\mathrm{/}9.81}\doteq 1.0096\text{ s } \\ t+t_3=t_2+t_1 \\ t=t_2+t_1-t_3=1.4347+1.0194-1.0096=1.4445\text{ s} \end{gathered}$$

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