Wie rechne ich den horizontalen bzw. schiefen Wurf aus?

S = Vektor Weg

S0 = Startvektor

V = Geschwingigkeitsvektor

G = Gravitation Vektor beachte G = (0,-g) da nach unten gerichtet

Weg:

Schiefer Wurf:

S = S0 + V*t + G*t²/2

s(t) = (x0,y0) + (cos(α)*|v|,sin(α)*|v|)*t + (0,-g)*t²/2

s(t) = (x0,y0) + (cos(α)*|v|*t,sin(α)*|v|*t) + (0,-gt²/2)

s(t) = (x0+cos(α)*|v|*t,y0+sin(α)*|v|*t-gt²/2)

also

sx = x0+cos(α)*|v|*t

sy = y0+sin(α)*|v|*t-gt²/2

horizontaler Wurf α = 0

S=S0 + V*t + G*t²/2

s(t) = (x0,y0) + (1*|v|,0*|v|)*t + (0,-g)*t²/2

s(t) = (x0,y0) + (1*|v|*t,0*|v|*t) + (0,-gt²/2)

s(t) = (x0,y0) + (|v|*t,0) + (0,-gt²/2)

s(t) = (x0+|v|*t,y0-gt²/2)

also

sx = x0+|v|

sy = y0-gt²/2

Geschwindigkeit v(t) entsprechend

v(t) = (cos(α)*|v|,sin(α)*|v|) + (0,-g*t)

v(t) = (cos(α)*|v|,sin(α)*|v|-g*t)

vx = cos(α)*|v|

vy = sin(α)*|v|-g*t

für α=0 horizontaler Wurf

v(t) = (cos(α)*|v|,sin(α)*|v|-g*t)

v(t) = (1*|v|,0*|v|-g*t)

vx = |v|

vy = -g*t