Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation

Already signed up?
New to KöMaL?

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.

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

103 students sent a solution.

3 points: 64 students.

2 points: 29 students.

1 point: 10 students.