At the opening of a 152 m deep pit, we throw a body vertically upwards with an initial speed of 29 m/s. How many seconds later should we drop another body from the same place into the pit so that both bodies reach the bottom of the pit simultaneously ( \( g=9.8 \) m/s\(^2\)).
The first body reaches the opening of the pit again after $\frac{2 c}{g} \mathrm{mp}$; it reaches the bottom of the pit after $\frac{-c+\sqrt{c^{2}+2 g s}}{g} \mathrm{mp}$. Therefore, the total travel time of the first body is

$$
\frac{2 c}{g}+\frac{-c+\sqrt{c^{2}+2 g s}}{g}=\frac{c+\sqrt{c^{2}+2 g s}}{g} .
$$

The travel time of the second body is $\sqrt{\frac{2 s}{g}}$, so the required time is:

$$
\frac{c+\sqrt{c^{2}+2 g s}}{g}-\sqrt{\frac{2 s}{g}}
$$

Substituting the given values, we find that the second body must be thrown into the trench $3.69 \mathrm{mp}$ after the first, so that both bodies reach the bottom of the trench simultaneously.

(Rássy Paulin. Eger.)

The problem was also solved by: Bartók I., Biró A., Braun I., Deusch E., Deutsch I., Enyedi B., Haar A., Hirschfeld Gy., Kertész G., Kürti I., Liebner A., Messer P., Neidenbach E., Pám M., Pazsiczky G., Pichler S., Pivnyik I., Popoviciu M., Prékopa D., Raab R., Ragány B., Riesz K., Riesz M., Rosenberg J., Schlesinger O., Schöffer I., Schwemmer I., Sonnenfeld J., Söpkéz Gy., Szávay Z., Szúcs A., Weisz P.