Mathematical and Physical Journal
for High Schools
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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.

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


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

Problems in Physics of KöMaL, February 2013