 Mathematical and Physical Journal
for High Schools
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# Problem P. 4541. (May 2013)

P. 4541. How long is the shadow of a 1 m long rod, which is fixed perpendicularly to the ground at the equator,

a) at noon, on the 21-st of June;

b) 2 hours later after noon on the 21-st of June?

(5 pont)

Deadline expired on June 10, 2013.

Sorry, the solution is available only in Hungarian. Google translation

Megoldásvázlat. $\displaystyle a)$ $\displaystyle h_1=\tg 23{,}45^\circ=0{,}43~\rm m.$

$\displaystyle b)$ $\displaystyle h_2=\sqrt{{\cos^{-2} 23{,}45^\circ\cdot \cos^{-2} 30^\circ}-1 }=0{,}76~\rm m.$

### Statistics:

 65 students sent a solution. 5 points: Bingler Arnold, Fehér Zsombor, Fekete Panna, Janzer Barnabás, Juhász Péter, Kollarics Sándor, Medek Ákos, Papp Roland, Reitz Angéla, Sal Kristóf, Szabó 928 Attila, Sztilkovics Milán. 4 points: Alabér Ádám, Mezősi Máté. 3 points: 19 students. 2 points: 21 students. 1 point: 2 students. 0 point: 9 students.

Problems in Physics of KöMaL, May 2013