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Uitgewerkte oefeningen fysica projectielbeweging
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Uitgewerkte oplossingen van de oefeningen die meegegeven werden bij de studiebegeleiding ivm fysica projectielbeweging
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KINEMATICA 2D - PROJECTIELBEWEGINGy
OEFENING 1
gegevens v0 = 15 m/s v
α = 70°
a = -g 70°
x
g
door de keuze van het assenkruis heeft de valversnelling g geen x-component
(A) baanvergelijkingen bij constante versnelling (B) v = vx²+ vy²
x = x0 + v0x.t + 1/2.ax.t² en tgα = vy/vx
y = y0 + v0y.t + 1/2.ay.t² vx = v0x + ax.t
t = 2s => x = 0 + 15.cos70.2 + 0 vy = v0y + ay.t
x = 10,26 m t = 2s => vx = 15.cos70 + 0 => tgα = vy/vx
y = 0 + 15.sin70.2 + 1/2.(-9,81).2² vx = 5,13 m/s tgα = -5,52/5,13
y = 8,57 m vy = 15.sin70 + (-9,81).2 tgα = -1,08
vy = -5,52 m/s => tg-1(vy/vx) = α
=> v = 5,13² + (-5,52)² = 7,54 m/s tg-1(-1,08) = -47,20°
(C) op maximale hoogte is vy = 0 (D) op maximale reikwijdte is y = 0
=> vy = v0y + ay.t => y = y0 + v0y.t + 1/2.ay.t²
0 = 15.sin70 + (-9,81).t 0 = 0 + 15.sin70.t + 1/2.(-9,81).t²
t = 15.sin,81 0 = 15.sin70 + 1/2.(-9,81).t
t= 1,44 s = tijd nodig om top te bereiken t = 15.sin70.,81
=> maximale hoogte na t = 1,44s => t = 2,87 s = tijd nodig om terug op grond te vallen
ymax = y0 + v0y.t + 1/2.ay.t² xmax = x0 + v0x.t + 1/2.ax.t²
ymax = 0 + 15.sin70.1,44 + 1/2.(-9,81).1,44² xmax = 0 + 15.cos70.2,87 + 0
ymax = 10,13 m xmax = 14,72 m
, OEFENING 2
we kiezen als nulpunt voor het assenkruis de uitgang van het kanon
(vanaf dat punt worden de meeste afmetingen gegeven)
door de keuze van het assenkruis heeft de valversnelling geen x-comp.
gegevens v0 = 26,5 m/s
θ0 = 53°
xtoren 1 = 23 m
htoren 1 = 18 m, door keuze assenkruis is y toren 1 = 18 - 3 = 15 m
hkanon = 3 m, door keuze assenkruis is ykanon = 3 - 3 = 0 m
hnet = 3 m, door keuze assenkruis is ynet = 3 - 3 = 0 m g
(A) op maximale reikwijdte is y = 0 (B) op maximale hoogte is vy = 0
y = y0 + v0y.t + 1/2.ay.t² vy = v0y + ay.t
0 = 0 + 26,5.sin53.t + 1/2.(-9,81).t² 0 = 26,5.sin53 + (-9,81).t
0 = 26,5.sin53 + 1/2.(-9,81).t t = 26,5.sin,81
t = -26,5.sin53.2 / (-9,81) t= 2,16 s = tijd nodig om top te bereiken
t = 4,31 s = tijd nodig om op net te landen maximale hoogte na t = 2,16s
xmax = x0 + v0x.t + 1/2.ax.t² ymax = y0 + v0y.t + 1/2.ay.t²
xmax = 0 + 26,5.cos53.4,31 + 0 ymax = 0 + 26,5.sin53.2,16 + 1/2.(-9,81).2,16²
xmax = 68,81 m ymax = 22,83 m tov hoogte uitgang kanon (keuze assenkruis)
dus ymax = 22,83 + 3 = 25,83 m boven grondniveau
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