De horizontale worp

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Gebaseerd op cursus SPC 2021

Verbeteringen/fouten/onduidelijkheden/...?

Deze open-source cursus is in ontwikkeling. De aanbevelingen van leerlingen om dit materiaal te verbeteren zijn erg welkom via info@wiskunde.opmaat.org
Dit kan gaan over:

Een voorbeeld dat onduidelijk is.
Onnauwkeurigheden, schrijffouten, ...
Een tussenstap die beter uitgelegd moet worden.
Een uitleg die je op youtube, wikipedia of aan de kersttafel gevonden hebt die ophelderend was.
...

Wanneer een object horizontaal met een bepaalde beginsnelheid wordt gekatapulteerd, noemen we die beweging een horizontale worp.

Wij beschouwen de worp in het luchtledige.

[Picture]

Figuur 1: De snelheid in horizontale richting verandert niet, die in de verticale richting neemt lineair toe in de tijd

In de beschrijving kunnen we de $x$-as horizontaal en de $y$-as verticaal naar beneden nemen. Omdat er volgens de $x$-as geen versnelling is het lichaam volgens de $y$-as valt met de valversnelling $g$, kunnen we de formules voor een ERB en een EVRB op de afzonderlijke assen toepassen en zo de volledige beweging beschrijven.

\begin{equation*} \left \{ \begin{array}{l} a_x=0\\ a_y=g \end{array} \right . \Rightarrow \left \{ \begin{array}{l} v_x=v_0\\ v_y=gt \end{array} \right . \Rightarrow \left \{ \begin{array}{l} x=v_0t\\ y=\frac {1}{2}gt^2 \end{array} \right . \end{equation*}

De baanvergelijking vinden we zoals eerder vermeld, door $t$ in functie van $x$ te schrijven $x=v_0t\Leftrightarrow t=\frac {x}{v_0}$ en dit in $y(t)$ te substitueren:

\begin{eqnarray*} y=\frac {1}{2}gt^2=\frac {1}{2}g\left (\frac {x}{v_0}\right )^2=\frac {g}{2v_0^2}x^2 \end{eqnarray*}

De baan is dus (een stuk van) een parabool.

Een vliegtuig vliegt met een snelheid van $450$ $\mathrm {km}/\mathrm {h}$ op een hoogte van $920$ $\mathrm {m}$.

Hoever voor het doel moeten de voedselpakketten gelost worden om op het doel terecht te komen?

De afstand waarover de voedselpakketten in horizontale richting zijn vooruit gegaan, kunnen we vinden met de baanvergelijking. We weten namelijk hoever de pakketten naar beneden zijn gevallen en wat hun beginsnelheid is:

\begin{eqnarray*} y&=&\frac {g}{2v_0^2}x^2\\ &\Downarrow &\\ x&=&v_0\sqrt {\frac {2y}{g}}=\SI {1712}{m} \end{eqnarray*}

Hoeveel tijd hebben de pakketten nodig om het doel te bereiken?

De valtijd voor de pakketten vinden we o.a. door naar de verticale valbeweging te kijken. Deze gebeurt onafhankelijk van wat er in de horizontale richting gebeurt, zodat:

\begin{eqnarray*} y&=&\frac {1}{2}gt^2\\ &\Downarrow &\\ t&=&\sqrt {\frac {2y}{g}}=\SI {13,7}{s} \end{eqnarray*}