Introduction to 2Dimensional Motion. 2Dimensional Motion. Definition: motion that occurs with both x and y components. Example: Playing pool . Throwing a ball to another person. Each dimension of the motion can obey different equations of motion. Solving 2D Problems.
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v = 40 m/s
v = 5 m/s, Q = 40o,
t = 2.5 min = 150 s
vx= v cosQvy= v sin Q
vx= 5 cos 40vy= 5 sin 40
vx= vy=
x = vxt y = vyt
x = ( )(150) y = ( )(150)
x = y =
vx
vy
v = 5 m/s
A particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2.
A particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2.
Vo,y = 6.2 m/s, Vo,x = 0 m/s, t = 5 s, ax = 4.4 m/s2, ay = 0 m/s2
x = ? y = ?
x = vo,x + at y = vo,y + at
x = 0 + (4.4)(5) y = 6.2 + (0)(5)
x = y =
A particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2.
A particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2.
vx = vo,x+ axtvy = voy + axt
vx = 0+ (4.4)(5)vy = 6.2+ 0(5)
vx = vy =
yo = 108 m, y = 0 m, g = 9.8 m/s2, vo,x = 3.6 m/s
v = ?
yo = 108 m, y = 0 m, g = 9.8 m/s2, vo,x = 3.6 m/s
v = ?
Gravity doesn’t change horizontal velocity. vo,x = vx =3.6 m/s
Vy2 = Vo,y2  2g(y – yo)
Vy2 = (0)2 – 2(9.8)(0 – 108)
Vy =
yo = 108 m, y = 0 m, g = 9.8 m/s2, vo,x = 3.6 m/s
v = ?
Gravity doesn’t change horizontal velocity. vo,x = vx =3.6 m/s
Vy2 = Vo,y2  2g(y – yo)
v = Vy2 = (0)2 – 2(9.8)(0 – 108)
Vy =
vo,x = 6.75 m/s, x = 8.95 m, y = 0 m, yo = 1.2, Vo,y = 0 m/s
g = ?
y = yo + Vo,yt  1/2gt2
g = 2(y  yo  Vo,yt)/t2
g = 2[0 – 1.2 – (0)( )]/( )2
g =
x = vo,xt
t = x/vo,x
t = 8.95/6.75
t =
Should be able to do this on your own!
General launch angle
General launch angleVo,x = Vo cos
Resolving the velocityVo
Vo,x = Vo cos
Resolving the velocityVo
vo = 9.5 m/s, q = 25o, g = 9.8 m/s2, Remember: because it lands at the same height: Dy = y – yo = 0 m and vy = vo,y
Find: Vo,y = Vo sin and Vo,x = Vo cos
Vo,y = 9.5 sin 25 Vo,x = Vo cos
Vo,y = Vo,x =
t = ?
Vy = Vo,y  gt
t = (Vy  Vo,y )/g
t = [( ) – ( )]/9.8 don’t forget vy = vo,y
t =
x
Trajectory of a 2D Projectilex
Trajectory of a 2D Projectilex
Range of a 2D ProjectileRange
x
Maximum height of a projectileMaximum
Height
Range
x
Maximum height of a projectileMaximum
Height
Range
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Maximum height of a projectileMaximum
Height
Range
g
g
g
g
g
x
Acceleration of a projectilex
Velocity of a projectilevx
vx
vy
vy
vx
vy
vx
vy
vx
x
Velocity of a projectilevx
vx
vy
vy
vx
vy
vx
vy
vx
x
Velocity of a projectilevx
vx
vy
vy
vx
vy
vx
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vx

vo
Velocity of a projectileThe purpose is to collect data to plot a trajectory for a projectile launched horizontally, and to calculate the launch velocity of the projectile. Equipment is provided, you figure out how to use it.
A golfer tees off on level ground, giving the ball an initial speed of 42.0 m/s and an initial direction of 35o above the horizontal.
How far from the golfer does the ball land?
Vo = 42m/s, q = 35o, g = 9.8 m/s2
R = ?