College Physics 1
Motion in Two Dimensions
• Reminder of vectors and vector algebra
• Displacement and position in 2-D
• Average and instantaneous velocity in 2-D
• Average and instantaneous acceleration in 2-D
✓ Projectile motion
Projectile Motion
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2-D problem and define a coordinate
system: x- horizontal, y- vertical (up +)
Try to pick x0 = 0, y0 = 0 at t = 0
Horizontal motion + Vertical motion
Horizontal: ax = 0 , constant velocity motion
Vertical:
ay = -g = -9.8 m/s2, v0y = 0
Equations:
Horizontal
Vertical
vx = v0 x + axt
v y = v0 y + a y t
x − x0 = v0 xt + 12 axt 2
y − y0 = v0 y t + 12 a y t 2
v x = v0 x + 2a x ( x − x0 ) v y 2 = v0 y 2 + 2a y ( y − y0 )
2
2
y f = yi + viy t − 12 gt 2
Projectile Motion
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X and Y motions happen independently, so
we can treat them separately
v x = v0 x
v y = v0 y − gt
x = x0 + v0 xt
y = y0 + v0 y t − 12 gt 2
Horizontal
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Vertical
Try to pick x0 = 0, y0 = 0 at t = 0
Horizontal motion + Vertical motion
Horizontal: ax = 0 , constant velocity motion
Vertical:
ay = -g = -9.8 m/s2
x and y are connected by time t
y(x) is a parabola
Projectile Motion
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2-D problem and define a coordinate system.
Horizontal: ax = 0 and vertical: ay = -g.
Try to pick x0 = 0, y0 = 0 at t = 0.
Velocity initial conditions:
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v0 can have x, y components.
v0x is constant usually. v0 x = v0 cos 0
v0y changes continuously. v0 x = v0 sin  0
Equations:
Horizontal
Vertical
v x = v0 x
v y = v0 y − gt
x = x0 + v0 xt
y = y0 + v0 y t − 12 gt 2
Trajectory of Projectile Motion
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Initial conditions (t = 0): x0 = 0, y0 = 0
v0x = v0 cosθ0 and v0y = v0 sinθ0
Horizontal motion:
x = 0 + v0 xt
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
t=
Vertical motion:
x
v0 x
y = 0 + v0 y t − 12 gt 2
 x  g x 
 − 

y = v0 y 
 v0 x  2  v0 x 
y = x tan  0 −
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2
g
2
x
2
2v0 cos2  0
Parabola;
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θ0 = 0 and θ0 = 90 ?
What is R and h ?
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Initial conditions (t = 0): x0 = 0, y0 = 0
v0x = v0 cosθ0 and v0x = v0 sinθ0, then
0 = 0 + v0 y t − 12 gt 2
x = 0 + v0 x t
t=
2v0 y
g
=
h
2v0 sin  0
g
2v cos 0 v0 sin  0 v0 sin 2 0
R = x − x0 = v0 x t = 0
=
g
g
2
t gt
h = y − y0 = v0 y th − 12 gth = v0 y −  
2 2 2
2
v0 sin 2  0
h=
2g
2
2
v y = v0 y − gt = v0 y − g
2v0 y
g
= −v0 y
Horizontal
Vertical
v x = v0 x
v y = v0 y − gt
x = x0 + v0 xt
y = y0 + v0 y t − 12 gt 2
Projectile Motion
at Various Initial Angles
• Complementary values
of the initial angle result
in the same range
• The heights will be
different
• The maximum range
occurs at a projection
angle of 45o
v0 sin 2
R=
g
2