1. Projectile Motion Practice
Launch at an Angle
This topic can be found on pp of your book.
“Sacred Rain Arrow” (1998)
Allan Houser

2. Practice Problem #1
Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go?
First you will need to draw a picture. Then, because the initial velocity is at an angle, we need to resolve the velocity vector into its x and y components.
We will then need to organize our knowns and unknowns.

3. Practice Problem #1
Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go?
x component
cosθ = 𝑎𝑑𝑗 ℎ𝑦𝑝
cos 42.5° = 𝑣 𝑥 𝑚/𝑠
vx = 20.8 m/s
y component
sinθ = 𝑜𝑝𝑝 ℎ𝑦𝑝
sin 42.5° = 𝑣 𝑦 𝑚/𝑠
viy = 19.1 m/s
28.2 m/s
19.1 m/s
42.5o
20.8 m/s

4. Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m (why?)
Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go?
x-direction
y-direction
Δx = ?
Δy = 0m (why?)
t = ?
v = 20.8 m/s
vi = 19.1 m/s
vf = m/s (why?)
a = m/s2
28.2 m/s
19.1 m/s
42.5o
20.8 m/s

5. Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m t = ?
Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go?
x-direction
y-direction
Δx = ?
Δy = 0m
t = ?
v = 20.8 m/s
vi = 19.1 m/s
vf = m/s
a = m/s2
Since there is not enough information in the x-direction to solve for Δx, we will need to solve for t in the y-direction first and then use that to solve for Δx.

6. Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m t = ?
Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go?
x-direction
y-direction
Δx = ?
Δy = 0m
t = ?
v = 20.8 m/s
vi = 19.1 m/s
vf = m/s
a = m/s2
vf = vi + at
-19.1m/s = 19.1m/s +
(-9.81m/s2)t
t = 3.89 s

7. Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m t = ?
Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go?
x-direction
y-direction
Δx = ?
Δy = 0m
t = ?
t = 3.89s
v = 20.8 m/s
vi = 19.1 m/s
vf = m/s
a = m/s2
vf = vi + at
-19.1m/s = 19.1m/s +
(-9.81m/s2)t
t = 3.89 s

8. Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m t = 3.89s
Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go?
x-direction
y-direction
Δx = ?
Δy = 0m
t = 3.89s
v = 20.8 m/s
vi = 19.1 m/s
vf = m/s
a = m/s2
vf = vi + at
-19.1m/s = 19.1m/s +
(-9.81m/s2)t
t = 3.89 s

9. Practice Problem #1 x-direction y-direction Δx = ? Δy = 0m t = 3.89s
Anita Knapp is a world class javelin thrower. If she throws a javelin with a velocity of 28.2 m/s at 42.5o, how far will the javelin go?
x-direction
y-direction
Δx = ?
Δy = 0m
t = 3.89s
v = 20.8 m/s
vi = 19.1 m/s
vf = m/s
a = m/s2
𝑣 𝑥 = ∆𝑥 ∆𝑡
20.8 𝑚/𝑠= ∆𝑥 3.89 𝑠
Δx = 80.9 m

10. Practice Problem #2
A cannon ball is shot off of the edge of a 85.0 m high cliff at an enemy ship. The ball has an initial velocity of 45.2 m/s at 38.1o. How far from the base of the cliff will the cannon ball hit?
Again we will first need to draw a diagram and then resolve the velocity vector into its components.

11. Practice Problem #2
A cannon ball is shot off of the edge of a 85.0 m high cliff at an enemy ship. The ball has an initial velocity of 45.2 m/s at 38.1o. How far from the base of the cliff will the cannon ball hit?
x component
cosθ = 𝑎𝑑𝑗 ℎ𝑦𝑝
cos 38.1° = 𝑣 𝑥 𝑚/𝑠
vx = 35.6 m/s
y component
sinθ = 𝑜𝑝𝑝 ℎ𝑦𝑝
sin 38.1° = 𝑣 𝑦 𝑚/𝑠
viy = 27.9 m/s
45.2 m/s
27.9 m/s
38.1o
35.6 m/s

12. Practice Problem #2 x-direction y-direction Δx = ? Δy = -85.0m t = ?
A cannon ball is shot off of the edge of a 85.0 m high cliff at an enemy ship. The ball has an initial velocity of 45.2 m/s at 38.1o. How far from the base of the cliff will the cannon ball hit?
x-direction
y-direction
Δx = ?
Δy = -85.0m
t = ?
v = 35.6 m/s
vi = 27.9 m/s
vf = ? m/s
a = m/s2
45.2 m/s
27.9 m/s
38.1o
35.6 m/s

13. Practice Problem #2
A cannon ball is shot off of the edge of a 85.0 m high cliff at an enemy ship. The ball has an initial velocity of 45.2 m/s at 38.1o. How far from the base of the cliff will the cannon ball hit?
∆𝑦= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2
-85.0m=(27.9m/s)t
+ ½(-9.81m/s2)t2
t = 7.89 s
x-direction
y-direction
Δx = ?
Δy = -85.0m
t = ?
v = 35.6 m/s
vi = 27.9 m/s
vf = ? m/s
a = m/s2

14. Practice Problem #2 𝑣 𝑥 = ∆𝑥 ∆𝑡 35.6 m/s = ∆𝑥 7.89 𝑠 Δx = 281 m
A cannon ball is shot off of the edge of a 85.0 m high cliff at an enemy ship. The ball has an initial velocity of 45.2 m/s at 38.1o. How far from the base of the cliff will the cannon ball hit?
𝑣 𝑥 = ∆𝑥 ∆𝑡
35.6 m/s = ∆𝑥 7.89 𝑠
Δx = 281 m
x-direction
y-direction
Δx = ?
Δy = -85.0m
t = 7.89
t = 7.89 s
v = 35.6 m/s
vi = 27.9 m/s
vf = ? m/s
a = m/s2

15. Practice Problem #3 Answer: 5.34 m above the ground.
Stan Still is standing 16.3 m away from a wall. If he throws a water balloon with a velocity of 15.2 m/s at 61.2o at the wall, how high on the wall will the water balloon hit?
Answer: 5.34 m above the ground.

16. y-direction equations
Projectile Motion Equations Summary
x-direction equation
y-direction equations
𝑣= ∆𝑥 ∆𝑡
𝑣 𝑓 = 𝑣 𝑖 +𝑎∆𝑡
∆𝑥= 1 2 (𝑣 𝑖 + 𝑣 𝑓 )∆𝑡
∆𝑥= 𝑣 𝑖 ∆𝑡+ 1 2 𝑎 ∆𝑡 2
∆𝑥= 𝑣 𝑓 ∆𝑡− 1 2 𝑎 ∆𝑡 2
𝑣 𝑓 2= 𝑣 𝑖 2+2𝑎∆𝑥

17. Important Points
vx is always constant in projectile motion constant!!!
You must resolve the initial velocity into its x- and y- components.
Acceleration is only due to gravity (down) and is always the same value: m/s2. The initial direction and angle of the projectile is irrelevant.
Time is the only variable that must be the same in both the x and y directions and is therefore the link between the two directions.