1. Projectile Motion ?
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2. First, let’s talk about Mr Morris’s boat….
                                                                                                       
If the boat has a speed of 10 m/s and the current is
5 m/s, what is the resultant velocity of the boat?
How long to travel across the river?

3. √10 2 + 52 = If the boat has a speed of 10 m/s and the current is
5 m/s, what is the resultant velocity of the boat? VR
5 m/s downstream Θ
10 m/s across
VR =
√ =
11.18 m/s
Θ =
Tan-1 (5/10) =
26.57º
downstream

4. How long to travel across the 120 m wide river?
The time to cross depends on the speed across the river.
t = d v
= 120 m
10m/s
= 12 sec
How far downstream will the boat land on the far bank?
The distance downstream depends on the downstream current speed and the time in the water.
d = vt
= (5 m/s)(12sec) = 60 m downstream

5. The BIG Rule; The perpendicular components of motion
are INDEPENDENT of each other
So… the velocity across the river is independent
of the velocity down the river.
We will use this rule again and again…

6. A PROJECTILE is an object that moves through space acted upon only by the earth’s gravity.
Dropped
Thrown up
Launched at an angle…

7. facts about projectiles;
1. Projectiles maintain a constant horizontal velocity (neglecting air resistance)
2. Projectiles always experience a constant vertical acceleration of “g” or 10 m/s2 (neglecting air resistance)

8. 3. Horizontal and vertical motion are completely INDEPENDENT of each other.
So we can separate the velocity of a projectile into horizontal and vertical components.

9. Let’s apply these rules to a horizontally thrown projectile…
A ball is thrown horizontally from a 100 m building at a velocity of 20 m/s. How far from the base of the building will it land?
For vertical;
dy = 100 m
V0 = 0 m/s
g = 10 m/s2
For horizontal;
dx = ?
Vx = 20 m/s
a = 0
We can use the vertical to find time…
100m
Since horizontal distance …
dy = Vot + ½at2
dx = Vxt
dx = (20m/s) ?
(4.47s)
100m = 0 + ½(10m/s2)t2
dx = 89.4 m
t = 4.47 s

10. = 3.13 s dy = vy0t + ½ gt2 = (0)(3.13) + ½ (10)(3.13)2 = 48.98 m
Example A person standing on a cliff throws a stone with a horizontal velocity of 15.0 m/s and the stone hits the ground
47 m from the base of the cliff. How high is the cliff?
vx = 15 m/s
dx = 47 m
vyo = 0
= 3.13 s
dy = vy0t + ½ gt2
= (0)(3.13) + ½ (10)(3.13)2
= m

11. A plane flying at 115 m/s drops a package from 600m
A plane flying at 115 m/s drops a package from 600m. How far from the drop point will it land?
Rule 5. Objects dropped from a moving vehicle have the same velocity as the moving vehicle.

12. Horizontal:
Vx = 115 m/s
dx = ?
Vertical:
Voy = 0
dy = 600 m
a = 10 m/s2
This is the same problem we’ve been working…
dy = ½ at2
dx = (115 m/s)(10.95s)
600 m= ½ (10m/s2)t2
dx = m
t = s

13. One more… A soccer ball is kicked horizontally off a 22
One more… A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
Horizontal Vertical
d = Vot + ½at2
Vx = ?
dx = 35.0m
Voy = 0
dy = 22.0m
a = 10 m/s2
22 m = 0 + ½(10m/s2)t2
t = 2.10 s
t = ?
t = ?
Vx = dx t
= m
2.10s
Vx = m/s

14. What is the vertical velocity just at impact? (Vyf)
Horizontal Vertical
Voy = 0
dy = 22.0m
a = 10 m/s2
Vfy = ?
Vx = 16.67m/s
dx = 35.0m
t = 2.10 s
Vfy = Vo + at
Vfy = 0 + (10m/s2)(2.10s)
Vfy = 21.0 m/s
What is the resultant velocity of the package at impact?
VR2 = (16.67m/s)2 + (21.0 m/s)2
= m/s VR
Vfy
Tan θ = 21 m/s
16.67 m/s
= 51.56º to the ground θ Vx

15. 15

16. facts about projectiles;
1. Projectiles maintain a constant horizontal velocity (neglecting air resistance)
2. Projectiles always experience a constant vertical acceleration of “g” or 10 m/s2 (neglecting air resistance)
3. Horizontal and vertical motion are completely INDEPENDENT of each other.
4. If the start height and the end height are the same, then time up equals the time down and the distance up equals the distance down
5. Objects dropped from a moving vehicle have the same velocity as the moving vehicle. 16

17. PROJECTILE MOTION AT AN ANGLE
The more general case of projectile motion occurs when the projectile is fired at an angle. 17

18. vox = 100 cos 30 = 86.60 m/s voy = 100 sin 30 = 50 m/s dx = vox t
Example: A golf ball is hit with an initial velocity of 100 m/s at an angle of 30 above the horizontal.
Find: a. Its position and velocity after 8 s
Let’s first find the components of the velocity
vox = 100 cos 30
= m/s
voy = 100 sin 30
= 50 m/s VR
Vyo
30º Vx
dx = vox t
= 86.60(8)
= m
dy = voy t + ½ gt2
= 50(8) + ½ (-10)(8)2
= m
vx = vox = m/s
vy = voy + gt
= 50 + (-10)(8)
= - 30 m/s 18

19. At top vfy = 0 vfy = voy + gt t = 5.0 s Total time T = 2t = 2(5.0)
b. The time required to reach its maximum height
At top vfy = 0 vfy = voy + gt
0 = 50 m/s + -10m/s2 t
t = 5.0 s
c. The horizontal range dx
Total time T = 2t
= 2(5.0)
= 10.0 s
x = vox t
= 86.6(10.0)
= 866 m 19

20. The angle that a projectile is launched will determine the range and maximum height it obtains.
Vy = VR sin θ
Vx = VR cos θ 20

21. Maximum range is an angle of 47º21

22. Need time… which trip to use ?
Example A grasshopper jumps between flowers 1.20 m apart. The grasshopper jumps to a height of 0.40 m, at what velocity and angle does he jump?
Horizontal Vertical
Need time… which trip to use ?
Vyo =
dy = .40 m
g = 10 m/s2
Vx =
dx = 1.2 m
t =
We will use the down trip.
0.57 s
dy = Vot + ½ at2
t = 0.28 s
0.40 m = 0 + ½ (10 m/s2)t2
tdown = 0.28 sec VR
ttotal = 0.57 sec Vy θ Vx 22

23. Vfy = Voy + at Vfy = Voy + 10 m/s2( 0.28s) Vfy = 2.80 m/s
Horizontal Vertical
Vyo =
dy = .40 m
g = 10 m/s2
Vx =
dx = 1.2 m
t =
Vfy = Voy + 10 m/s2( 0.28s)
0.57 s
t = 0.28 s
Vfy = 2.80 m/s
VR2 = (2.80 m/s)2 + (2.11 m/s)2
VR = m/s
Vx = dx t
= 1.20 m
0.57s
Tan θ = 2.80 m/s
2.11 m/s
Vx = 2.11 m/s
Θ = 53.00º above
The ground VR Vy θ Vx 23

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