PROJECTILE MOTION
When we throw a ball :
•  There is a constant velocity horizontal
motion
•  And there is an accelerated vertical
motion
•  These components act
independently of each other
PROJECTILE MOTION
•  A falling object with constant linear velocity
and vertical acceleration :
2-D motion
•  The path or trajectory projectiles make is
parabolic (neglecting air resistance).
•  Two independent motions- horizontal and
vertical.
•  Use kinematics equations in one direction
at a time.
•  The connection between the two motions is
the variable time.
Projectile motion
•  Vertical motion is free fallconstant acceleration motion.
becomes…
Δx = vi t + ( ½) a t 2
vf = vi + at
vf2 = vi2 + 2aΔx
Projectile motion
•  Horizontal motion is constant
velocity motion.
..and ax= 0, so Δx = vi t + ( ½) a t2
becomes
Problem Type 1:
•  A projectile is launched
with an initial horizontal
velocity from an elevated
position and follows a
parabolic path to the
ground. Predictable
unknowns include the
initial speed of the
projectile, the initial height
of the projectile, the time
of flight, and the
horizontal distance of the
projectile.
Problem Type 2:
•  A projectile is launched at an
angle to the horizontal and
rises upwards to a peak while
moving horizontally. Upon
reaching the peak, the
projectile falls with a motion
which is symmetrical to its path
upwards to the peak.
Predictable unknowns include
the time of flight, the horizontal
range, and the height of the
projectile when it is at its peak.
Type 1 examples
•
A pool ball leaves a 0.60-meter high
table with an initial horizontal velocity of
2.4 m/s. Predict the time required for the
pool ball to fall to the ground and the
horizontal distance between the table's
edge and the ball's landing location.
A pool ball leaves a 0.60-meter high table with an initial horizontal
velocity of 2.4 m/s. Predict the time required for the pool ball to fall to
the ground and the horizontal distance between the table's edge and
the ball's landing location.
List Givens
•  Horizontal (x) info:
•  Vertical (y) info:
•  x = ?
•  vix = 2.4 m/s
•  ax = 0 m/s2
•  y = -0.60 m
•  viy = 0 m/s
•  ay = -9.8 m/s2
t=?
Equations:
Δx = vi t + ( ½) a t2
vf = vi + at
vf2 = vi2 + 2aΔx
Choose an equation and solve
•  y = viy t + ( ½) ay t2
•
•
•
•
-0.60 m = (0 m/s)•t + 0.5•(-9.8 m/s/s)•t2
-0.60 m = (-4.9 m/s/s)•t2
0.122 s2 = t2
t = 0.350 s
Now use equation in horizontal
direction to solve for x
•
•
•
•
x = vixt + 0.5axt2
x = (2.4 m/s)(0.3499 s) + 0.5•(0)•(0.3499 s)2
x = (2.4 m/s)•(0.3499 s)
x = 0.84 m
Example B, problem type 1
•  A soccer ball is kicked horizontally off a
22.0-meter high hill and lands a distance
of 35.0 meters from the edge of the hill.
Determine the initial horizontal velocity of
the soccer ball.
A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a
distance of 35.0 meters from the edge of the hill. Determine the initial
horizontal velocity of the soccer ball.
List Givens
•  Horizontal (x) info:
•  Vertical (y) info:
•  x = 35 m
•  vix = ?
•  ax = 0 m/s2
•  y = -22.0 m
•  viy = 0 m/s
•  ay = -9.8 m/s2
Δx = vi t + ( ½) a t2
vf = vi + at
vf2 = vi2 + 2aΔx
Solve
•  y = viyt + 0.5ayt2
•  t= 2.12 s
•  Now use x = vixt + 0.5axt2 to solve for vix
•  vix = 16.5 m/s
Previous Examples = Problem Type 1
Problem Type 2
Projectiles launched
at an angle
•  initial velocity = vi
•  launch angle θ
•  separate initial velocity into
components
vi
vyi
θ
vx
Components from trig. func.
•  Use sine and cosine functions to
find components.
Problem type 2
(example A)
•
A football is kicked with an initial velocity
of 25 m/s at an angle of 45-degrees with
the horizontal. Determine the time of
flight, the horizontal distance, and the
peak height of the football.
back to the problem…
A football is kicked with an initial velocity of 25 m/s at an angle of
45o with the horizontal. Determine the time of flight, the
horizontal distance, and the peak height of the football.
•  HORIZONTAL COMPONENT
•  VERTICAL COMPONENT
•  vix = vi • cosθ
•  viy = vi • sinθ
•  vix = 25 m/s • cos 45o •  viy = 25 m/s • sin 45o
•  vix = 17.7 m/s
•  viy = 17.7 m/s
A football is kicked with an initial velocity of 25 m/s at an angle of
45o with the horizontal. Determine the time of flight, the
horizontal distance, and the peak height of the football.
List Givens
•  Horizontal (x) info:
•  Vertical (y) info:
•
•
•
•
•
•
•
•
x = ???
vix = 17.7 m/s
vfx = 17.7 m/s
ax = 0 m/s2
Equations:
y = ???
viy = 17.7 m/s
vfy = -17.7 m/s **
ay = -9.8 m/s2
Δx = vi t + ( ½) a t2
vf = vi + at
vf2 = vi2 + 2aΔx
*symmetry of projectile motion
Same concept of symmetry applies to free fall motion
Use vertical info to find t
vfy = viy + ayt
-17.7 m/s = 17.7 m/s + (-9.8 m/s2) t
-35.4 m/s = (-9.8 m/s2) t
3.6077 s = t = 3.61 s
Now find horizontal distance
•  x = vix t + ½ axt2
•  x = (17.7 m/s)(3.6077 s) + ½ (0 m/s/s)(3.6077 s)2
•  x = (17.7 m/s)•(3.6077 s)
•  x = 63.8 m
Find “height of projectile at its peak”
(in other words, the vertical displacement, y)
•  Note: The height is reached when at a time= ½ t
•  y = viyt + ½ ayt2
•  y = (17.7 m/s)(1.80 s) + ½ (-9.8 m/s/s)(1.80 s)2
•  y = 31.9 m + (-15.9 m)
•  y = 15.9 m
Find “height of projectile at its peak”
(in other words, the vertical displacement, y)
•  Note: You could also use vf2 = vi2 + 2aΔy as long as
you know vfy = 0 at the peak height
Type 2: example B
•  A long jumper leaves the ground with an
initial velocity of 12 m/s at an angle of 28o
above the horizontal. Determine the time
of flight, the horizontal distance, and the
peak height of the long-jumper.
A long jumper leaves the ground with an initial velocity of 12 m/s at an
angle of 28o above the horizontal. Determine the time of flight, the
horizontal distance, and the peak height of the long-jumper.
•  HORIZONTAL COMPONENT
•  VERTICAL COMPONENT
•  vix = vi • cosθ
•  viy = vi • sinθ
•  vix = 12 m/s • cos 28o •  viy = 12 m/s • sin 28o
•  vix = 10.6 m/s
•  viy = 5.6 m/s
A long jumper leaves the ground with an initial velocity of 12 m/s at an
angle of 28-degrees above the horizontal. Determine the time of flight,
the horizontal distance, and the peak height of the long-jumper.
List Givens
•  Horizontal (x) info:
•  Vertical (y) info:
•
•
•
•
•
•
•
•
•
x = ???
vix = 10.6 m/s
vfx = 10.6 m/s
ax = 0 m/s2
Equations:
ypeak = ???
viy = 5.6 m/s
vfy = -5.6 m/s
vypeak = 0 m/s
ay = -9.8 m/s2
Δx = vi t + ( ½) a t2
vf = vi + at
vf2 = vi2 + 2aΔx
Use vertical info to find t
•  vfy = viy + ayt
•  -5.6 m/s = 5.6 m/s + (-9.8 m/s2) t
•  -11.2 m/s = (-9.8 m/s2) t
•  1.1497 s = t = 1.1 s
Now find horizontal distance
•  x = vix t + ½ axt2
•  x = (10.6 m/s)(1.1497 s) + ½ (0 m/s/s)(1.1497 s)2
•  x = (10.6 m/s)•(1.1497 s)
•  x = 12.2 m
Find “height of projectile at its peak”
(in other words, the vertical displacement, y)
•  Note: The height is reached when at a time= ½ t
•  y = viyt + ½ ayt2
•  y = (5.6 m/s)(0.5748 s) + ½ (-9.8 m/s2)(0.5748 s)2
•  y = 3.219 m + (-1.6189 m)
•  y = 1.6 m