Question

From the top of a tower first body thrown up and reaches the ground in 9 s. Second body thrown down with the same initial speed and reaches ground in 4 s, then time taken by freely falling third body to reach to the ground from same height is

Answer 1

Displacement of a first body (thrown up)

$$d=-v_it_1+\frac{gt_1^2}{2}$$

Displacement of a second body (thrown down)

$$d=v_it_2+\frac{gt_2^2}{2}$$

So

$$-v_it_1+\frac{gt_1^2}{2}=v_it_2+\frac{gt_2^2}{2}$$

Therefore, the initial velocity

$$v_i=\frac{g}{2}(t_1-t_2)=\frac{10}{2}(9-4)=25\:\rm{m/s}$$

The height of a tower

$$d=-25\times 9+\frac{10\times 9^2}{2}=180\:\rm{m}$$

Finally, time taken by freely falling third body

$$t_3=\sqrt{\frac{2d}{g}}=\sqrt{\frac{2\times 180}{10}}=6\:\rm{s}$$