Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem P. 4509. (February 2013)

P. 4509. 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 March;

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

(3 pont)

Deadline expired on March 11, 2013.

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

Megoldásvázlat. $\displaystyle a)$ Nincs árnyéka a botnak.

$\displaystyle b)$ $\displaystyle \frac{1}{\sqrt{3}}\cdot1~{\rm m} = \tg 30^\circ\cdot1~{\rm m}\approx 58~\rm cm.$

### Statistics:

 104 students sent a solution. 3 points: 65 students. 2 points: 29 students. 1 point: 10 students.

Problems in Physics of KöMaL, February 2013