1. Projectile Motion

2. A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is an upside down parabola.
Figure Caption: This strobe photograph of a ball making a series of bounces shows the characteristic “parabolic” path of projectile motion.

3.  Components of motion can
Projectile Motion
Projectile Motion  Motion of an object that is projected into the air at an angle.
Near the Earth’s surface, the acceleration a on the projectile is downward and equal to a = g = 9.8 m/s2
Goal: Describe projectile motion after it starts.
Galileo: Analyzed horizontal & vertical components of motion separately.
Today: Displacement D & velocity v are vectors
 Components of motion can
be treated separately

4. Projectile Motion  vx = vx0 , x = vx0t
Simplest example: A ball rolls across a table, to the edge & falls off the edge to the floor. It leaves the table at time t = 0. Analyze the y part of motion & the x part of motion separately.
y part of the motion: Down is positive & the origin is at table top: y0 = 0. Initially, there is no y component of velocity: vy0 = 0
 vy = gt, y = (½)g t2
x part of motion: The origin is at the table top: x0 = 0.
No x component of acceleration(!) ax = 0.
Initially the x component of velocity is vx0
 vx = vx0 , x = vx0t

5. A Ball Rolls Across Table & Falls Off
Projectiles can be understood by analyzing horizontal vertical motions separately. At any point, v has both x & y components. Take down as positive. Initial velocity has an x component ONLY! That is vy0 = 0. The kinematic equations tell us that, at time t,
vx = vx0, vy = gt
x = vx0t
y = vy0t + (½)gt2
t = 0 here
Figure Caption: Projectile motion of a small ball projected horizontally. The dashed black line represents the path of the object. The velocity vector at each point is in the direction of motion and thus is tangent to the path. The velocity vectors are green arrows, and velocity components are dashed. (A vertically falling object starting at the same point is shown at the left for comparison; vy is the same for the falling object and the projectile.)

6. vx = vx0 , vy = gt Summary A ball rolling across the table & falling.
The vector velocity v has 2 components:
vx = vx0 , vy = gt
Vector displacement D has 2 components:
x = vx0t , y = (½)g t2

7. positions of the 2 balls are identical at identical times,
The speed in the x direction is
constant; in the y-direction the
object moves with constant
acceleration g.
The photo shows 2 balls that
start to fall at the same time.
The one on the right has an
initial speed in the x direction.
It can be seen that the vertical
positions of the 2 balls are
identical at identical times,
while the horizontal position
of the yellow ball increases
linearly.
Figure Caption: Multiple-exposure photograph showing positions of two balls at equal time intervals. One ball was dropped from rest at the same time the other was projected horizontally outward. The vertical position of each ball is seen to be the same at each instant.

8. The SAME as free fall motion!!
Projectile Motion
PHYSICS
The Vertical (y) part of the motion:
vy = gt , y = (½)g t2
The SAME as free fall motion!!
 An object projected horizontally will reach the ground at the same time as an object dropped vertically from the same point!
(x & y motions are independent)

9. A common example of a projectile is a baseball!
Projectile Motion
A common example of a projectile is a baseball!

10. A “Somewhat General” Case Components of initial velocity v0:
An object is launched at initial angle θ0 with the horizontal. Analysis of the motion is similar to before, except the initial velocity has a vertical component vy0  0. Let up be positive now!
Components of initial velocity v0:
vx0 = v0cosθ0 vy0 = v0sinθ0
The Parabolic shape of the path is real.
Figure Caption: Path of a projectile fired with initial velocity v0 at angle θ0 to the horizontal. Path is shown dashed in black, the velocity vectors are green arrows, and velocity components are dashed. The acceleration a = dv/dt is downward. That is, a = g = -gj where j is the unit vector in the positive y direction.
Acceleration = g down
for the entire trip!

11. No Acceleration in the x Direction!
General Case: Take y positive upward & origin at the point where it is shot: x0 = y0 = 0
vx0 = v0cosθ0, vy0 = v0sinθ0
Horizontal Motion:
No Acceleration in the x Direction!
vx = vx0 , x = vx0 t
Vertical Motion:
vy = vy0 - gt , y = vy0 t - (½)g t2
(vy) 2 = (vy0)2 - 2gy
If y is positive downward, the - signs become + signs.

12. Summary: Projectile Motion
Motion with constant acceleration in 2 dimensions, where the acceleration is g and is down.

13. Solving Projectile Motion Problems
Read the problem carefully, & choose the object(s) you are going to analyze.
Sketch a diagram.
Choose an origin & a coordinate system.
Decide on the time interval; this is the same in both directions, & includes only the time the object is moving with constant acceleration g.
Solve for the x and y motions separately.
List known & unknown quantities. Remember that vx never changes, & that vy = 0 at the highest point.
Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

14. Projectile Motion Example 4.4: Rolling off a Cliff
A car rolls off a cliff of height
h = 20 m. Its initial velocity is
v0 = 10 m/s, along the horizontal.
Calculate
a. The time it takes to hit the
ground at the base after it leaves the cliff.
b. Its horizontal distance x from the base when it hits the ground.
c. Its velocity when it hits the ground.

15. Example: Driving off a cliff!!
A movie stunt driver on a motorcycle speeds horizontally off a 50 m high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90 m from the base of the cliff where the cameras are?
vx = vx0 = ? vy = -gt
x = vx0t, y = - (½)gt2
Time to bottom:
t = √2y/(-g) = 3.19 s
vx0 = (x/t) = 28.2 m/s
y is positive upward, y0 = 0 at top. Also vy0 = 0

16. Example: Kicked Football
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A football is kicked at an angle θ0 = 37.0° with a velocity of m/s, as shown. Calculate: a. Max height. b. Time when hits ground. c. Total distance traveled in the x direction. d. Velocity at top. e. Acceleration at top.
θ0 = 37º, v0 = 20 m/s
 vx0= v0cos(θ0) = 16 m/s, vy0= v0sin(θ0) = 12 m/s