We throw a heavy object vertically upwards with an initial velocity of $120 \mathrm{~m}$; 5 seconds later, we throw another object with the same initial velocity. After how many seconds will the two objects meet; $g=9.8$.

 untranslated portion: $g=9.8$.
At the moment of meeting, the speed of the two bodies is equal; thus:

$$
v_{1}=v_{2}
$$

If we denote the initial speed by $c$, then the time of ascent for the first body is:

$$
t=\frac{c}{g}
$$

Let $t_{1}$ be the time during which the first body travels from its highest point to the meeting point, so the second body ascends for $t-5+t_{1}$ seconds before meeting the first body. Considering (1), we get:

$$
g t_{1}=c-g\left(\frac{c}{g}-5+t_{1}\right)
$$

from which

$$
t_{1}=2.5 \mathrm{~s}
$$

The two bodies meet $t+t_{1}=14.7 \mathrm{~s}$ after the first body is launched.

The problem was solved by: Friedmann Bernát, S.-A.-Ujhely; Galter János, Sz.-Udvarhely; Suschnik József, Kecskemét; Szabó Gusztáv, Győr; Hofbauer Ervin and Pósch Gyula, Budapest; Grünhut Béla and Visnya Aladár, Pécs.