Projectile Motion

1. •Speed- rate at w/c an object travels
in a certain distance.
•Velocity- speed of an object in a
given direction/ speed plus
direction.
•Accelaration- the rate of change in
velocity

2. •Direction: Answer the following questions and write
your solution in 1/2 crosswise.
1. What is the velocity of car that traveled a total of 75
km North in 1.5 hours?
2. An ant carries food at a speed of 1 m/s. How long will
it take the ant to carry a cookie crumb from the kitchen
table to the ant hill, a distance of 5000 m? Express your
answers in seconds, minutes and hours.
3. A train moves from rest to a speed of 25 m/s in 30
seconds. What is its acceleration?

3. •the acceleration of an object remains
constant irrespective of time.
1. A ball rolling down an incline.
2. A car moving with a constant acceleration.
3. An object is dropped from a certain height.
4. A ball is thrown vertically straight and goes
down.
Uniformly Accelerated Motion (UAM)

4. •the acceleration of an object
remains constant irrespective of
time.
3. An object is dropped from a
certain height.
Uniformly Accelerated Motion (UAM)

5. 0m/s
-9.81m/s
-19.62m/s
-29.43m/s
-39.24m/s
0s
1s
2s
3s
4s
Note: All
object fall
at the
same rate,
if there is
no air
resistance
regardless
of its
mass.

6. •the acceleration of an object
remains constant irrespective of
time.
4. A ball is thrown vertically straight
and goes down.
Uniformly Accelerated Motion (UAM)

7. 0m/s
9.81m/s
19.62m/s
29.43m/s
3s
2s
1s
0s
0m/s 0s
-9.81m/s 1s
-19.62m/s 2s
-29.43m/s 3s

8. Uniformly Accelerated Motion (UAM)
•4 UAM Equation
Formulas
1. Vf= Vi+at
2. d= Vit+1/2at2
3. Vf
2
=Vi
2
+2ad
4.d= (Vf-Vi/2)t
•Variables
Vf- final velocity -m/s
Vi- initial velocity -m/s
a- acceleration -m/s2
t- time -s
d- displacement -m

9. Uniformly Accelerated Motion (UAM)
•4 UAM Equation
Formulas
1. Vf= Vi+at
2. d= Vit+1/2at2
3. Vf
2
=Vi
2
+2ad
4.d= (Vf-Vi/2)t
Problem
1. From rest a car
accelerated at 8m/s2
for
10s.
a. What is the position of
the car at the end of the
10s?
b. What is the velocity of
the car at the end of the
10s?

10. a.
Given:
a= 8m/s2
Vi= 0m/s
t= 10s d=?
Solution:
d= Vit+1/2at2
d= (0m/s)(10s)+(0.5)(8m/s2
)(10s)2
d= (4m/s2
)(100s2
)
d= 400 m
b.
Given:
a= 8m/s2
Vi= 0m/s
t= 10s Vf=?
Solution:
Vf= Vi+at
Vf= 0m/s+(8m/s)2
(10s)
Vf= 80m/s

11. Uniformly Accelerated Motion (UAM)
•4 UAM Equation
Formulas
(Horizontal)
1. Vfx= Vix+axt
2. dx= Vixt+1/2axt2
3. Vfx
2
=Vix
2
+2axdx
4.dx= (Vfx-Vix/2)t
2. With an initial
velocity of 20m/s, a car
accelerated at 8m/s2
for 10 seconds.
a. What is the position
of the car after 10
seconds.

12. a.
Given:
ax= 8m/s2
Vix= 20m/s
t= 10s dx=?
Solution:
•4 UAM Equation
Formulas
(Horizontal)
1. Vfx= Vix+at
2. dx= Vixt+1/2at2
3. Vfx
2
=Vix
2
+2adx
4.dx= (Vfx-Vix/2)t

13. •Free Fall- when an object is acted upon
by the sole influence of gravity.
ay= - 9.81m/s2
1. An object is dropped.
2. An object is thrown vertically straight
and goes back down.
Uniformly Accelerated Motion (UAM)

14. 0m/s
-9.81m/s
-19.62m/s
-29.43m/s
-39.24m/s
0s
1s
2s
3s
4s
Note: All
object fall
at the
same rate,
if there is
no air
resistance
regardless
of its
mass.

15. 0m/s
9.81m/s
19.62m/s
29.43m/s
3s
2s
1s
0s
0m/s 0s
-9.81m/s 1s
-19.62m/s 2s
-29.43m/s 3s

16. Uniformly Accelerated Motion (UAM)
•Variables
Vf- final velocity -m/s
Vi- initial velocity -m/s
a- acceleration -m/s2
t- time -s
d- displacement -m
•4 UAM Equation
Formulas (Vertical)
1. Vfy= Viy+ayt
2. dy= Viyt+1/2ayt2
3. Vfy
2
=Viy
2
+2aydy
4.dy= (Vfy-Viy/2)t

17. Uniformly Accelerated Motion (UAM)
Problem:
1. A rock was dropped on
top of the building and hits
the ground after 5 seconds.
a. What is the height of the
building?
Given:
t=5s ay=-9.81m/s2
Viy=0m/s d=?
•4 UAM Equation
Formulas (Vertical)
1. Vfy= Viy+ayt
2. dy= Viyt+1/2ayt2
3. Vfy
2
=Viy
2
+2aydy
4.dy= (Vfy-Viy/2)t

18. Uniformly Accelerated Motion (UAM)
Problem:
1.
Given:
t=2s ay=-9.81m/s2
Viy=? Vfy=0m/s
•4 UAM Equation
Formulas (Vertical)
1. Vfy= Viy+ayt
2. dy= Viyt+1/2ayt2
3. Vfy
2
=Viy
2
+2aydy
4.dy= (Vfy-Viy/2)t

20. Projectile Motion
•any object that moves in two dimensions and forms
a curved path and is acted upon by gravity.
Projectile- object being launced. It has independent
horizontal and vertical velocities.
Trajectory- pattern of a projectile through space
called Parabola.
Range- Max. horizontal distance traveled by a
projectile.

25. Vertical(UAM)
1. Vfy= Viy+ayt
2. dy= Viyt+1/2ayt2
3. Vfy
2
=Viy
2
+2aydy
4.dy= (Vfy-Viy/2)t
•Angle Launch:
Horizontal(V=d/t)
1. dx=VXt
ax=0m/s
•Angle Launch:
Vix=Vcos
Projectile Motion

26. VX= 80 m/s
dy=- 50 m
dx=?
a. t=?
b. dx=?
1.

27. Vertical(UAM)
Given:
dy=-50m
ay=-9.81m/s2
viy=0m/s
Horizontal(V=d/t)
Given:
Vx=80m/s
dx=?
t=?
Projectile Motion

28. Projectile Motion
2. A ball was kicked at an angle of 450
upward at 80
m/s on a cliff that is 50m.

29. Special Equation for Projectile Motion
•R(range)= Vi
2
sin20/g
•H(height)=Vi
2
(sin0)2
/2g
•t(A-B)=Visin0/g
•t(A-C)=2Visin0/g

30. Assignment: Solve the following in a one whole sheet
of paper. Show the given, formula and solution.
1. An arrow is fired directly horizontal off a cliff that is 10.0m tall with a
velocity of 65.5m/s.
a. How long is the arrow in the air?
b. What is the range of the arrow?
2. A rock is thrown with a velocity of 23.7m/s horizontally off the top of an
elevated hill. If the time it takes to reach the fround is 5.70 seconds. What
is the height of the hill?
3. A football is kicked at ground level with a speed of 20 m/s at a 40 degree
angle to the horizontal. How much later does it hit the ground? What’s the
range of the football?