Mathematical and Physical Journal
for High Schools
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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