How long does it take for a freely falling body to fall so that in the subsequent 1.3 seconds it falls $49.34 \mathrm{~m}$? 1[^0]


[^0]:    ${ }^{1}$ The acceleration due to free fall: $g=980.8 \frac{\mathrm{cm}}{\mathrm{sec}^{2}}$ (according to Gruber's determination, in Budapest $g=980.837 \frac{\mathrm{cm}}{\mathrm{sec}^{2}}$).
Let the sought time be denoted by $t$, then

$$
\begin{aligned}
& \frac{g}{2}(t+1.3)^{2}=\frac{g}{2} t^{2}+49.34 \\
& t=\frac{49.34}{1.3 g}-\frac{1.3}{2}=3.22 \text{ sec}
\end{aligned}
$$

(Freund Ernó, Budapest.)

The problem was also solved by: Ádámffy E., Bánó L., Blum J., Csada I., Dévai I., Dömény I., Ehrenstein P., Erdős V., Fodor H., Földes R., Fuchs I., Haar A., Harsányi Z., Jánosy Gy., Kiss J., Krampera Gy., Kräuter F., Kürti I., Láng O., Martini I., Merse P., Morvai O., Pető L., Rássy P., Schwarz O., Székely J., Tandlich E.