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 let the second body fall freely from the same height to hit the Earth simultaneously?

Correct answer:

t =  1.4445 s

Step-by-step explanation:

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

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