A juggler throws two balls vertically upwards at intervals of 0.7 seconds, with an initial velocity of $14.56 \frac{\mathrm{m}}{\mathrm{sec}}$. After how much time and at what height will the balls meet, and with what speed will they collide: 1 .[^0]


[^0]:    ${ }^{1}$ The acceleration due to gravity: $g=980.8 \frac{\mathrm{cm}}{\mathrm{sec}^{2}}$ (according to Gruber's determination, in Budapest $g=980.837 \frac{\mathrm{cm}}{\mathrm{sec}^{2}}$ ).
Let the sought time be $t$ sec from the first ball being thrown, then the path of the first ball until the meeting is $c t-\frac{g}{2} t^{2}$,

the second " " " " $c(t-m)-\frac{g}{2}(t-m)^{2}$,

where $c$ is the initial speed, and $m$ represents $0.7 \mathrm{sec}$.

According to the problem,

$$
c t-\frac{g}{2} t^{2}=c(t-m)-\frac{g}{2}(t-m)^{2},
$$

from which

$$
t=\frac{c}{g}+\frac{m}{2}=1.84 \mathrm{sec}
$$

and

$$
\begin{gathered}
s=c t-\frac{g}{2} t^{2}=10.17 \mathrm{~m} \\
v=c-g t=3.43 \frac{m}{\mathrm{sec}}
\end{gathered}
$$

The speed of the other ball is the same, but with the opposite sign.

(Jánosy Gyula, Budapest.)

The problem was also solved by: Ádámffy E., Blum J., Csada I., Fodor H., Földes R., Fuchs I., Haar A., Kiss J., Krampera Gy., Kräuter F., Kürti I., Merse P., Pető L., Rássy P., Rosenberg J., Schwarz O., Singer D., Székely J., Tandlich E.