What is Projectile Motion?
Instructional Objectives:
• Students will be able to:
– Define Projectile Motion
– Distinguish between the different types of
projectile motion
– Apply the concept to a ball rolling off a table and
measure its velocity
Projectile Motion
• Same requirements as free fall, but projectile
motion is two-dimensional motion of an object
– Vertical direction
– Horizontal direction
Concepts of Projectile Motion
• Horizontal
– Motion of a ball rolling freely along a
level surface
– Horizontal velocity is ALWAYS constant
– Nothing pushes or pulls on it in the
horizontal direction to cause it to
accelerate
• Vertical
– Motion of a freely falling object
– Force due to gravity
– Vertical component of velocity changes
with time
• Parabolic
– Path traced by an object accelerating only
in the vertical direction while moving at
constant horizontal velocity
Parabola
Examples of Projectile Motion
• Launching a Cannon ball
Factors Affecting Projectile
Motion
• What two factors would affect projectile
motion?
– Angle
– Initial velocity
Initial Velocity
Angle

The range R is the horizontal distance the projectile
has traveled when it returns to its launch height.
Projectile Motion Solutions
PROBLEM-SOLVING STRATEGY Projectile motion problems
MODEL Make simplifying assumptions.
VISUALIZE Use a pictorial representation. Establish a coordinate system with the x-axis horizontal and the yaxis vertical. Show important points in the motion on a sketch. Define symbols and identify what the problem is
trying to find.
SOLVE The acceleration is known:
ax  0 and a y  - g , thus the problem becomes one of kinematics.
The
kinematic equations are (THESE ARE NOT NEW – THEY ARE JUST OUR 1-D EQUATIONS):
xf  xi  vix t
vfx  vix  constant
yf  yi  viy t  12 g (t )2
vfy  viy  g t
t is the same for the horizontal and vertical components of the motion. Find t from one component, then use
that value for the other component.
ASSESS Check that your result has the correct units, is reasonable, and answers the question.
Horizontal motion
x  x 0  v0 x t
x  x 0  ( v 0 cos 0 ) t
• No acceleration
Vertical motion (Equations of
Motion ):
1)
2)
3)
1 2
y  y0  v0 y t  g t
2
1 2,
 ( v 0 sin 0 ) t  g t
2
v y  v 0 sin 0  g t
v y  ( v 0 sin 0 )  2 g ( y  y 0 )
2
2
For a particular range less than the maximum
and for a particular launch velocity,
two different launch angles will give that range.
The two angles add to give 900.
450 gives the maximum range.
Example Problem 1
An object is fired from the ground at 100 meters
per second at an angle of 30 degrees with the
horizontal
a) Calculate the horizontal and vertical
components of the initial velocity
b) How long does it take to reach highest point?
c) How far does the object travel in the horizontal
direction?
Example: A Home Run
A baseball is hit so that it leaves the
bat making a 300 angle with the
ground. It crosses a low fence at the
boundary of the ballpark 100 m from
home plate at the same height that it
was struck. (Neglect air resistance.)
What was its velocity as it left the
bat?
v0 x  v0 cos  ; v0 y  v0 sin  ;
x1  x0  v0 x (t1  t0 )  (v0 cos  )t1 ;
y1  0  y0  v0 y (t1  t0 )  12 g (t1  t0 ) 2  (v0 sin  )t1  12 gt12 ;
0  (v0 sin  )t1  12 gt12  (v0 sin   12 gt1 )t1;
so t1  0 or t1  2v0 sin  / g ;
Therefore, x1  (v0 cos  )t1  (v0 cos  )(2v0 sin  / g );
x1  2v0 2 sin  cos  / g  v0 2 sin 2 / g ;
v0  x1 g / sin 2  (100 m)(9.80 m/s 2 ) / sin(60)  33.6 m/s
Sample Problem 2
• In Fig. 4-15, a rescue
plane flies at 198 km/h
(= 55.0 m/s) and a
constant elevation of
500 m toward a point
directly over a boating
accident victim
struggling in the water.
The pilot wants to
release a rescue capsule
so that it hits the water
very close to the victim.
(a) What should be the angle  of the pilot's line of sight to the
victim when the release is made?
Solution
  tan
1
x
h
x  x 0  ( v 0 cos 0 ) t
1 2
y  y 0  ( v 0 sin  0 ) t  g t
2
1
 500 m  (55.0 m / s) (sin 0 ) t  (9.8 m / s 2 ) t 2
2

Solving for t, we find t = ± 10.1 s (take the positive root).
x  0  (55.0 m / s) (cos 0 ) (10.1 s)
x  555.5 m
  tan
1
555.5 m
 48
500 m

As the capsule reaches the water, what is its velocity v
(b)
as a magnitude and an angle?
When the capsule reaches the water,
v x  v 0 cos  0  (55.0 m / s) (cos 0  )  55.0 m / s
v y  v0 sin 0  g t
 (55.0 m / s) (sin 0 )  (9.8 m / s 2 ) (10.1 s)
  99.0 m / s
v  113 m / s and
55 m/s
   61
61°
99 m/s
113 m/s
Sample Problem 3
Figure 4-17 illustrates the flight of Emanuel Zacchini over three Ferris
wheels,
 0 located as shown and each 18 m high. Zacchini is launched
with speed v0 = 26.5 m/s, at an angle  0 = 53° up from the horizontal
and with an initial height of 3.0 m above the ground. The net in which
he is to land is at the same height.
• (a) Does he clear the first Ferris wheel?
SOLUTION
• The equation of trajectory when x0 = 0 and y0 = 0 is given by :
g x2
y  ( tan  0 ) x 
2 ( v 0 cos  0 ) 2
Solving for y when x = 23m gives
( 9.8 m / s 2 ) ( 23 m 2 )
 ( tan 53 ) (23 m) 
2 ( 26.5 m / s) 2 (cos 53 ) 2
 20.3 m

Since he begins 3m off the ground, he clears the Ferris wheel
by (23.3 – 18) = 5.3 m
(b) If he reaches his maximum height when he is
over the middle Ferris wheel, what is his clearance
above it?
• SOLUTION:
At maximum height, vy is 0. Therefore,
v y  (v 0 sin 0 ) 2  2 gy  0
2
( v 0 sin  0 ) 2 (26.5 m / s) 2 (sin 53 ) 2
y

 22.9 m
2
2g
(2) (9.8 m / s )
and he clears the middle Ferris wheel by
(22.9 + 3.0 -18) m =7.9 m
(c) How far from the cannon should the center of
the net be positioned?
• SOLUTION:
2
2
v0
(26.5 m / s)

R
sin 2 0 
sin (2 * 53 )
2
g
9.8 m / s
 69 m
FAST-MOVING PROJECTILES - SATELLITES
• Let’s throw stones horizontally with ever
increasing velocity.
• The Earth’s curvature is 16 ft for every 5 miles
(4.9 m for 8 km).
Throw an object faster, faster, faster.
5 miles
16 ft
We draw in each trajectory for 1 second
Vescape = 7 mi/s = 11 km/s
Vcircle = 5 mi/s = 8 km/s
Ellipse
Ellipse
Ellipse - Circle
Ellipse
Ellipse
A rock is thrown upward at an angle.
What happens to the horizontal
component of its velocity as it rises?
(Neglect air resistance.) 33% 33% 33%
(a) it decreases
(b) it increases
(c) it remains the same
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In baseball which path would a home run
most closely approximate? (Neglect air
resistance.)
33%
33%
33%
(a) hyperbolic
(b) parabolic
(c) ellipse
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A projectile is launched upward at an angle of 75° from
the horizontal and strikes the ground a certain distance
down range. What other angle of launch at the same
launch speed would produce the same distance?
(Neglect air resistance.)
33%
33%
33%
(a) 45°
(b) 15°
(c) 25°
a
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A horizontally traveling car drives off of a cliff next
to the ocean. At the same time that the car leaves the
cliff a bystander drops his camera. Which hits the
ocean first? (Neglect air resistance.)
33%
33%
33%
(a) car
(b) camera
(c) they both hit at the
same time
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