Problem P. 5034. (May 2018)

P. 5034. How long did the object, projected horizontally at an initial speed of \(\displaystyle v_0\), fall while it reached a position which was at a distance of \(\displaystyle s\) from the position of the projection? (Neglect air resistance.)

Data: \(\displaystyle v_0=5\) m/s, \(\displaystyle s=20\) m.

(4 pont)

Deadline expired on June 11, 2018.

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

Megoldás. A \(\displaystyle v_0\) kezdősebességgel eldobott test elmozdulásvektorának hossza \(\displaystyle t\) idejű esés után:

\(\displaystyle s=\sqrt{\left(v_0t\right)^2+\left(\frac{g}2 t^2\right)^2}.\)

Innen \(\displaystyle t^2\)-re egy másodfokú egyenletet kaphatunk:

\(\displaystyle g^2(t^2)^2+4v_0^2(t^2)-4s^2=0,\)

aminek pozitív megoldása:

\(\displaystyle t^2=\frac{2v_0^2}{g^2}\left(\sqrt{1+\frac{g^2s^2}{v_0^4}}-1 \right), \)

vagyis az esés ideje

\(\displaystyle t=\frac{\sqrt{2}v_0}{g}\sqrt{ \sqrt{1+\frac{g^2s^2}{v_0^4}}-1 }=1{,}89 ~\rm s.\)

Ellenőrzés:

\(\displaystyle x=5~\frac{\rm m}{\rm s}\cdot (1{,}89~{\rm s})= 9{,}47~\rm m, \)

\(\displaystyle y=\frac{1}{2}\,9{,}81~\frac{\rm m}{\rm s^2}\cdot (1{,}89~{\rm s})^2=17{,}61~\rm m, \)

\(\displaystyle \sqrt{x^2+y^2}=20~{\rm m}.\)

Statistics:

67 students sent a solution.
4 points:	54 students.
3 points:	6 students.
2 points:	1 student.
1 point:	6 students.

Problems in Physics of KöMaL, May 2018