1. Angled Launch
Physics

2. Projectile: Any object that is in free fall. Examples: pumpkins
sky divers
arrows
rocks
vomit
etc…

3. Angled Launch L  Launch velocity v0 Vertical launch velocity
v0y = v0sin
Horizontal launch velocity
y- dimension:
v0y fights off gravity and gives the projectile air time.
x- dimension:
v0x keeps the projectile moving forward at a constant speed until it strikes something.
v0x = v0cos

4. Let’s analyze a soccer ball that has been kicked at an angle.
20 m/s
20 m/s
9.8 m/s
19.6 m/s
20 m/s
20 m/s
-9.8 m/s
29.4 m/s
20 m/s
20 m/s
-19.6 m/s
y – dimension loses 9.8 m/s every second
x – dimension continues unaffected
Notice the launch velocity = impact velocity
-29.4 m/s
20 m/s

5. SYMBOLIC KINEMATICS END ∆x = v0 = v = a = t =
Y – DIMENSION:
END ∆x = v0 = v = a = t =
TOP ∆x = v0 = v = a = t =
0 m
height
Lsin
Lsin
- Lsin
0 m/s
- 9.8 m/s2
- 9.8 m/s2
ttotal
tup = ½ ttotal

6. SYMBOLIC KINEMATICS
X – DIMENSION:
∆x = v = t =
distance
Lcos
ttotal

7. NUMERIC EXAMPLE: END ∆x = v0 = v = a = t =
Bob launches a rocket moving 150 m/s at an angle of degrees. How high will it go? Where will it land?
END
∆x =
v0 =
v =
a =
t =
Use Equation 1 to solve for time:
0 m
Lsin = 150 sin (35) = 86.0
v = at + v0
- Lsin = -150 sin (35) = -86.0
t = v - v0 a
= -86 – 86
-9.8
- 9.8 m/s2 =
ttotal = need

8. NUMERIC EXAMPLE: TOP Use Equation 3 to solve for ∆x : ∆x = v0 = want o
Bob launches a rocket moving 150 m/s at an angle of degrees. How high will it go? Where will it land?
TOP
∆x =
v0 =
v =
a =
t =
Use Equation 3 to solve for ∆x :
want o
Lsin = 150 sin (35) = 86.0
∆x = ½ (v0 +v)t
0 m/s
∆x = ½ (v0)t
= ½ (86)( )
- 9.8 m/s2
tup = /2 = =
378 m

9. NUMERIC EXAMPLE: ∆x = v = want t = Lcos = 150 cos (35) = 122.8728
Bob launches a rocket moving 150 m/s at an angle of degrees. How high will it go? Where will it land?
∆x =
v =
t =
want
Lcos = 150 cos (35) =
Use the velocity equation to solve for ∆x :
v = ∆x ∆t
∆x = v∆t
= ( )( ) =
2157 m

10. NUMERIC EXAMPLE:
How high and far will Bob’s rocket go if he changes the angle to 45 degrees?

11. NUMERIC EXAMPLE:
How high and far will Bob’s rocket go if he changes the angle to 55 degrees?

12. MAXIMUM DISTANCE & HEIGHT
Let’s make a table of what we learned. It turns out there are some useful patterns we have observed.
ANGLE
HEIGHT
DISTANCE 35 45 55
378 m
574 m
770 m
2157 m
2296 m

13. SAME IMPACT SITE:
MAX DISTANCE:
MAX HEIGHT:
If two launch angles are complementary (add up to 90 degrees) they will both land at the same location. (One goes high, one goes low, but they both hit the same target)
If you want to obtain the maximum distance, launch with a 45 degree angle.
If you want to obtain the maximum height, launch with a 90 degree angle.

14. Advanced Angled Launch
Physics

15. Advanced Angled Launch
What if you are looking for a point on the trajectory that is not located at the maximum height or distance? IF YOU KNOW THE TIME: Plug it in to both the x and y dimensions.

16. EXAMPLE: X ∆x = v = t = Y ∆x = v0 = v = a = t = want want 13 cos (40)
An angry bird is launched 13 m/s at an upward angle of 40 degrees. Where will the bird be half a second later? X
∆x =
v =
t = Y
∆x =
v0 =
v =
a =
t =
want
want
13 m/s
13 cos (40)
13 sin (40)
40 degrees
0.5 s
don’t care
- 9.8 m/s2
Image from:
v = ∆x ∆t
∆x = v∆t
0.5 s
= (13 cos (40))(0.5)
∆x = ½ at2 + v0t
∆x = ½ (-9.8)(0.5)2 + (13 sin (40)(0.5) =
4.98 m =
2.95 m

17. Advanced Angled Launch
IF YOU KNOW THE HORIZONTAL POSITION: Use the x-dimension to solve for time and then plug it in.

18. EXAMPLE: X ∆x = v = t = Y ∆x = v0 = v = a = t = 1.2 m want 8 cos (32)
Bob uses his spoon to catapult some mashed potatoes at the wall. The wall is 1.2 meters away from his chair. The potatoes leave his spoon moving 8 m/s at an upward angle of 32 degrees. Where on the wall will they strike? X
∆x =
v =
t = Y
∆x =
v0 =
v =
a =
t =
1.2 m
want
8 m/s
8 cos (32)
8 sin (32)
need
don’t care
32 degrees
- 9.8 m/s2
v = ∆x ∆t
∆t = ∆x/v
0.177 s
= 1.2/(8 cos (32))
∆x = ½ at2 + v0t
∆x = ½ (-9.8)(0.177)2 + (8 sin (32)(0.177) =
0.177 s =
0.597 m above the spoon

19. Advanced Angled Launch
IF YOU WANT TO KNOW WHETHER THE PROJECTILE WAS MOVING UPWARDS OR DOWNWARDS: Calculate the time UP and compare it with the current time.

20. EXAMPLE: Y ∆x = v0 = max height v = at + v0 v = a = 13 sin (40)
Was the angry bird from example 1 still moving up or was it coming back down? Y
∆x =
v0 =
v =
a =
t =
max height
v = at + v0
13 sin (40)
t = - v0 / a
0 m/s
= - (13 sin (40)) / (-9.8)
- 9.8 m/s2 =
0.853 s
time up
The bird can go up for seconds. It is currently only 0.5 seconds and therefore the bird is going UP.

21. EXAMPLE: Y ∆x = v0 = max height v = at + v0 v = a = 8 sin (32)
Were the potatoes from example 2 still moving up or were they coming back down? Y
∆x =
v0 =
v =
a =
t =
max height
v = at + v0
8 sin (32)
t = - v0 / a
0 m/s
= - (8 sin (32)) / (-9.8)
- 9.8 m/s2 =
0.433 s
time up
The potatoes can go up for seconds. It is currently only seconds and therefore the potatoes are still going UP.