1. Projectile Motion

2. Background Information
Projectile motion problems must be solved by breaking the movement of the projectile into its x (horizontal) and y (vertical) components
Projectiles are any objects that are thrown or launched into the air and are subject to gravity (g = -9.8 m/s2)

3. Projectiles (neglecting air resistance) follow a path called a parabola
The red arrows indicate movement in the y direction and blue arrows indicate movement in x direction y x y x y x y x x y

4. Today we are studying projectiles launched with an initial horizontal velocity from an elevated position
Equations used to solve the problems are our Acceleration equations:
a = ∆v/∆t
xf = xi +vit + ½at2
vf2 = vi2 + 2a∆x

5. When solving the problems, you must solve in terms of the x (horizontal) information and separately for the y (vertical) information

6. HINTS…
1. When calculating the x component, we neglect all air resistance. Therefore, acceleration in the x direction is
a = 0 m/s2 and this is true for ALL problems

7. HINTS…
2. When calculating the y component of acceleration we assume a = -9.8 m/s2 since the only force acting on the object is gravity. Also, since there is no initial velocity in the y direction, vi = 0 m/s (in y direction only)

8. Example
A billiards ball falls off a .60 m high table with an initial velocity of 2.4 m/s. Calculate the following:
a) the time required for the ball to fall to the ground
b) the horizontal distance between the table’s edge and the ball’s landing position.

9. X Y
xf = ?
yf = -.60 m
vi = 2.4 m/s
vi = 0 m/s
a = 0 m/s2
a = -9.8 m/s2
t = ?
Step 1: List known and unknown variables in columns labeled X and Y
yf is is negative to show that the displacement was downward…aka a negative direction

10. Step 2: Solve for the first unknown using the column that contains the most information
Hint: Usually you need to solve for time first
To solve for t: yf = yi +vit + ½at2
-.6 = 0 + 0t + ½(-9.8)t2
-.6 = -4.9t2
.123 = t2
.35 s = t

11. Step 3: Using the time, solve for any other unknowns
To solve for the horizontal distance:
xf = xi +vit + ½at2
xf = (.35) + .5(0)(.35)2
xf = .84 meters

12. Let’s Do One Together!
A soccer ball is kicked horizontally off a 22 m high hill and lands 35 m away from the edge of the hill.
Determine the initial horizontal velocity of the ball.

13. Answer
t = 2.11 s
vi = 16.6 m/s