1. Projectile Motion

2. Projectile Motion Objects that are thrown are called projectiles.
The path of a projectile is a curve called a parabola.

3. Projectile motion ∆ 𝑑 𝑦 = 𝑣 𝑖𝑦 𝑡− 1 2 𝑔 𝑡 2 ∆ 𝑑 𝑥 = 𝑣 𝑖𝑥 𝑡
A projectile’s motion can be broken down into the horizontal and vertical components:
∆ 𝑑 𝑦 = 𝑣 𝑖𝑦 𝑡− 1 2 𝑔 𝑡 2
∆ 𝑑 𝑥 = 𝑣 𝑖𝑥 𝑡
Key Concepts:
Time is the same in both the x and y
Acceleration is gravity m/s2
Down is negative
*Remember to break the initial velocity into its components!

4. Launched at an angle
Vertical and Horizontal components to initial velocity:
𝑣 𝑖𝑥 = 𝑣 𝑖 𝑐𝑜𝑠𝜃
𝑣 𝑖𝑦 = 𝑣 𝑖 𝑠𝑖𝑛𝜃
12 m/s
40°
𝑣 𝑖𝑥 =12𝑐𝑜𝑠40
𝑣 𝑖𝑦 =12𝑠𝑖𝑛40

5. Example
Mark drives his motorcycle off a horizontal ramp at 325 m/hr and lands a horizontal distance of 30 m away from the edge of the ramp. What is the height of the ramp?
Variables:
Δ 𝑑 𝑥 = 30 m Δ 𝑑 𝑦 = ?
a = m/s2 𝑣 𝑥 =325 𝑚/ℎ𝑟

6. Example ∆ 𝑑 𝑦 = 𝑣 𝑦 𝑡+ 1 2 𝑎 𝑡 2 ∆ 𝑑 𝑥 = 𝑣 𝑥 𝑡 Variables:
a = m/s2 𝑣 𝑥 =325 𝑚/ℎ𝑟
30 = (325)t
t = 30/325
t = 0.09 sec
∆ 𝑑 𝑦 = (0)(0.09) + ½(-9.81)( )
∆ 𝑑 𝑦 = m

7. Example
A baseball is thrown at an angle of 25° at a speed of 23 m/s. If the ball was caught 42 m from the thrower, how long was it in the air? How high did the ball travel?
Variables:
Θ = 25 t= ?
V = 23 𝑑 𝑦 = ?
𝑑 𝑥 = 42

8. Example ∆ 𝑑 𝑦 = 𝑣 𝑦 𝑡 − 1 2 𝑎 𝑡 2 ∆ 𝑑 𝑥 = 𝑣 𝑥 𝑡 Variables:
∆ 𝑑 𝑦 = 𝑣 𝑦 𝑡 − 1 2 𝑎 𝑡 2
∆ 𝑑 𝑥 = 𝑣 𝑥 𝑡
Example
Variables:
Θ = 25 t= ? 𝑣 𝑖𝑥 = 𝑣 𝑖 𝑐𝑜𝑠𝜃
V = 23 𝑑 𝑦 = ? 𝑣 𝑖𝑥 =23𝑐𝑜𝑠25
𝑑 𝑥 = 𝑣 𝑖𝑥 = 20.8
42 = (20.8)t
t = 42/20.8
t = 2.01sec
∆ 𝑑 𝑦 = 9.72(2.01) - ½(-9.81)( )
∆ 𝑑 𝑦 = m

9. Finding 𝑣 𝑖 𝑣 𝑖 = 2 −𝑔 𝑑 𝑥 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
𝑣 𝑖 = 2 −𝑔 𝑑 𝑥 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
What is the initial velocity of an object thrown a horizontal distance of 250 m at an angle of 30°?