A NCC parade is going at a uniform speed of 9 km h$$^{-1}$$ under a mango tree on which a monkey is sitting at a height of 19.6 m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is: (Given $$g = 9.8 \ m s^{-2}$$)

A monkey drops a mango from a height of 19.6 m and we need to determine how far from the tree the cadet who catches it is at the moment of the drop. Using the relation $$h = \frac{1}{2}gt^2$$, we set $$19.6 = \frac{1}{2} \times 9.8 \times t^2$$ which gives $$t^2 = \frac{19.6 \times 2}{9.8} = 4$$ and hence $$t = 2 \text{ s}$$.

The parade speed is 9 km/h, which converts to $$9 \times \frac{5}{18} = 2.5$$ m/s. In the 2 seconds before the mango reaches the ground, the cadet walks a distance $$d = v \times t = 2.5 \times 2 = 5 \text{ m}$$.

Therefore, the cadet must be 5 m away from the tree at the moment of the drop to arrive directly underneath when the mango falls. Hence, the correct answer is Option A: 5 m.