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 fall-

constant acceleration motion. becomes…

Δx = vi t + ( ½) a t 2

vf = vi + at

vf2 = vi

2 + 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.

List Givens

•  Horizontal (x) info:

•  x = ? •  vix = 2.4 m/s •  ax = 0 m/s2

Equations:

•  Vertical (y) info:

•  y = -0.60 m •  viy = 0 m/s •  ay = -9.8 m/s2

Δx = vi t + ( ½) a t2 vf = vi + at vf

2 = vi2 + 2aΔx

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.

t = ?

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.

List Givens •  Horizontal (x) info:

•  x = 35 m •  vix = ? •  ax = 0 m/s2

•  Vertical (y) info:

•  y = -22.0 m •  viy = 0 m/s •  ay = -9.8 m/s2

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.

Δx = vi t + ( ½) a t2 vf = vi + at vf

2 = 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

θ

vi vyi

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

•  vix = vi • cosθ •  vix = 25 m/s • cos 45o •  vix = 17.7 m/s

•  VERTICAL COMPONENT

•  viy = vi • sinθ •  viy = 25 m/s • sin 45o •  viy = 17.7 m/s

List Givens

•  Horizontal (x) info:

•  x = ??? •  vix = 17.7 m/s •  vfx = 17.7 m/s •  ax = 0 m/s2

Equations:

•  Vertical (y) info:

•  y = ??? •  viy = 17.7 m/s •  vfy = -17.7 m/s ** •  ay = -9.8 m/s2

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.

Δx = vi t + ( ½) a t2 vf = vi + at vf

2 = 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 = vi

2 + 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

•  vix = vi • cosθ •  vix = 12 m/s • cos 28o •  vix = 10.6 m/s

•  VERTICAL COMPONENT

•  viy = vi • sinθ •  viy = 12 m/s • sin 28o •  viy = 5.6 m/s

List Givens

•  Horizontal (x) info:

•  x = ??? •  vix = 10.6 m/s •  vfx = 10.6 m/s •  ax = 0 m/s2

Equations:

•  Vertical (y) info:

•  ypeak = ??? •  viy = 5.6 m/s •  vfy = -5.6 m/s •  vypeak = 0 m/s •  ay = -9.8 m/s2

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.

Δx = vi t + ( ½) a t2 vf = vi + at vf

2 = 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