Mathematical and Physical Journal
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Problem P. 4864. (October 2016)

P. 4864. A body was projected at two different angles of elevation at the same initial speed, such that the range of the projection was the same in both cases. What are these two angles if the time of the motion is \(\displaystyle n\) times as much in one of the cases than in the other? Air drag is to be neglected in both cases.

(4 pont)

Deadline expired on November 10, 2016.

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

Megoldás. Ha \(\displaystyle v_0\) kezdősebességgel, a vízszinteshez képest \(\displaystyle \alpha\) szögben hajítunk el egy testet, akkor a függőleges irányú kezdősebessége \(\displaystyle v_0\sin\alpha\), a mozgásának ideje

\(\displaystyle t=\frac{2v_0}{g}\sin\alpha.\)

Ennyi idő alatt a test vízszintes irányban

\(\displaystyle d=tv_0 \cos\alpha=\frac{v_0^2}{g}\sin(2\alpha)\)

távolságra jut el. Ha ez a \(\displaystyle d\) távolság két különböző elhajítási szögre (\(\displaystyle \alpha_1\)-re és \(\displaystyle \alpha_2\)-re) is ugyanakkora, akkor

\(\displaystyle \sin(2\alpha_1)=\sin(2\alpha_2), \qquad \text{vagyis}\qquad \alpha_2=90^\circ-\alpha_1.\)

A mozgásidők aránya:

\(\displaystyle n=\frac{t_1}{t_2}=\frac{\sin\alpha_1}{\sin\alpha_2}=\frac{\sin\alpha_1}{\sin(90^\circ-\alpha_1)}=\tg\alpha_1.\)

Ezek szerint az elhajítások szöge

\(\displaystyle \alpha_1=\arctg n\qquad \text{és} \qquad \alpha_2=90^\circ-\alpha_1=\arctg \frac{1}{n}.\)


113 students sent a solution.
4 points:67 students.
3 points:12 students.
2 points:11 students.
1 point:12 students.
0 point:11 students.

Problems in Physics of KöMaL, October 2016