Introduction to projectiles motion GCSE .pdf

projectiles motion GCSE

1. Projectile Motion
•
•
•
Projectiles are objects which are fired into the air.
Their motion obeys the kinematic equations if we
ignore air resistance.
The major difference is that a projectile's motion
occurs in two directions. (horizontal and vertical)

2. Projectile Motion
•
•
All projectile problems have two main parts, their
motion in the vertical (y) direction and their motion in
the horizontal (x) direction.
The motion in each of these directions is
independent of one another.

3. Projectile Motion
•Objects launched with a horizontal velocity and
dropped from the same height will always strike the
ground at the same time.

4. Projectile Motion
•
•
•
Why?
Because they both have the same height and an
initial vertical velocity of zero!
The time a projectile spends in the air depends upon
its vertical height and initial vertical velocity.

5. Solving Projectile Motion
•
•
•
1) Draw a diagram
2) Let up be positive and down
negative.
3) Break the velocity into the x-comp
and y-comp. Make a table stating
the givens in the x-direction and
y-direction.

6. Solving Projectile Motion
•
•
4) Solve for the time (could be in x or y
direction)
5) Once you solve for the time in one
direction you can use it to solve for
motion in the other direction

7. Sample Problem
•
•
Prarthna throws Rachel off a 50.0 m
high cliff with a horizontal velocity of
3.0 m/s. How far from the base will she
strike the ground?
Ans. dx = 9.58 m

8. Solution
•
•
•
•
•
•
You need at least 3 variables in the y-dir or 2
variables in x-dir to solve.
We have 3 variables in y direction so use
them to solve for time (t).
d = v
1
t + 0.5at
2
-50 = 0 + 0.5(-9.81)t
2
t
2
= 10.19
t = 3.19s
X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2
v = 3 m/sv = 3 m/s d = -50 md = -50 m
d = ?d = ? v
1 = 0v
1
= 0
t = ?t = ? t = ?t = ?

9. Solution
•
•
•
•
•
•
Once you solve for the time it can be used in
the y or x direction.
a = 0 in x-dir ALWAYS, so use v=d/t
v = d/t
d = vt
d = 3(3.19)
d = 9.58 m
X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2
v = 3 m/sv = 3 m/s d = -50 md = -50 m
d = ?d = ? v
1 = 0v
1
= 0
t = 3.19 st = 3.19 st = 3.19 st = 3.19 s

10. Sample Problem
•
•
Pierre hits a golf ball at 75 m/s at an angle of 12
0
above the horizontal.
Calculate:




a) The maximum height of the ball
b) the time the ball was in the air
c) how far it traveled in the horizontal direction before it
strikes the ground
d) The velocity when it strikes the ground
A = 12.4 m, 3.18s, 233.4m, 75m/s 12
0
below horzizontal

11. Solution
•
•
•
•
•
Break the velocity into its x and y
components.
v
x
= 75 cos 12
v
x
= 73.4 m/s
v
y
= 75 sin 12
v
y
= 15.6 m/s


X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2



v = 73.4 m/sv = 73.4 m/sv
1 = 15.6 m/sv
1
= 15.6 m/s
12
hyp
75 m/s
oppv
y
adjv
x

12. Part A
•
•
•
•
•
•
At max. height v = 0 ALWAYS!
Now we have enough data in y-dir. to solve for
d.
(v
2
)
2
= (v
1
)
2

+ 2ad
0 = (15.6)
2

+ 2(-9.81)d
d = -(15.6)
2
/ 2(-9.81)
d = 12.4 m
X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2
v = 73.4 m/sv = 73.4 m/sv
1 = 15.6 m/sv
1
= 15.6 m/s
d = ?d = ? v
2 = 0v
2
= 0
d = ?d = ?t = ?t = ?

13. Part B
•
•
•
•
•
To solve for time in the air we know when the ball
returns to the same height v
1
= -v
2
Now we have enough data to solve for t.
v
2
= v
1
+ at
t = (-15.6 – 15.6)/-9.81
t = 3.18 s

X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2
v = 73.4 m/sv = 73.4 m/sv
1 = 15.6 m/sv
1
= 15.6 m/s
d = ?d = ? v
2 = -15.6 m/sv
2
= -15.6 m/s
t = ?t = ?t = ?t = ?

14. Part C
•
•
•
•
•
Now that we know the time in the y-dir, we can use
it in the x-dir. to solve for d.
v = d/t
d = vt
d = 73.4(3.18)
d = 233.4 m

X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2
v = 73.4 m/sv = 73.4 m/sv
1 = 15.6 m/sv
1
= 15.6 m/s
d = ?d = ? v
2 = -15.6 m/sv
2
= -15.6 m/s
t = 3.18 st = 3.18 st = 3.18 st = 3.18 s

15. Part D
•
•
To solve for v
2
when the ball returns to the same
height we know v
1
= -v
2
Therefore v
2
= 75m/s 12
0
below horizontal

16. Assessment As Learning #5
•
•
Answer the following questions to the best of your
ability and give yourself a mark out of 10 based on
your number of correct answers.

This mark does not count towards your average, but
gives you feedback as to how well you are
understanding the concepts.

17. Assessment as Learning
•A golf ball is hit from a tee-off that is 10 m above
ground level at 75 m/s, at an angle of 12
0
above the
horizontal. Calculate:
a) The maximum height of the ball (above the ground)
b) the time the ball was in the air
c) how far it traveled in the horizontal direction before it
strikes the ground
d) The velocity when it strikes the
ground
A = 22.4 m, 3.73 s, 274 m,
76.3 m/s at 16
o
below horiz

18. Solution
•
•
•
•
•
Break the velocity into its x and y
components.
v
x
= 75 cos 12
v
x
= 73.4 m/s
v
y
= 75 sin 12
v
y
= 15.6 m/s


X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2



v = 73.4 m/sv = 73.4 m/sv
1 = 15.6 m/sv
1
= 15.6 m/s

19. Part A
•
•
•
•
In the previous problem we solved for d = 12.4
m.
We need to add 10 m to this value to get d in
this problem, since the ball was 10 m above the
ground when hit.
d = 12.4 + 10
d = 22.4 m

X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2
v = 73.4 m/sv = 73.4 m/sv
1 = 15.6 m/sv
1
= 15.6 m/s
d = ?d = ? v
2 = 0v
2
= 0
d = ?d = ?t = ?t = ?

20. Part B
•
•
•
•
•
To solve for time in the air we know d = -10
when the ball strikes the ground
Now we have enough data to solve for v
2
in the y
direction.
(v
2
)
2
= (v
1
)
2

+ 2ad
(v
2
)
2
= (15.6)
2

+ 2(-9.81)(-10)
v
2
= -21 m/s

X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2
v = 73.4 m/sv = 73.4 m/sv
1 = 15.6 m/sv
1
= 15.6 m/s
d = ?d = ? d = -10 md = -10 m
t = ?t = ?t = ?t = ?

21. Part B
•
•
•
•
Now we have enough data to solve for t.
v
2
= v
1
+ at
t = (-21 – 15.6)/-9.81
t = 3.73 s


X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2
v = 73.4 m/sv = 73.4 m/sv
1 = 15.6 m/sv
1
= 15.6 m/s
d = ?d = ? v
2 = -21 m/sv
2
= -21 m/s
d = -10 md = -10 mt = ?t = ?

22. Part C
•
•
•
•
•
Now that we know the time in the y-dir, we can use
it in the x-dir. to solve for d.
v = d/t
d = vt
d = 73.4(3.73)
d = 274 m

X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2
v = 73.4 m/sv = 73.4 m/sv
1 = 15.6 m/sv
1
= 15.6 m/s
d = ?d = ? v
2 = -21 m/sv
2
= -21 m/s
t = 3.73 st = 3.73 st = 3.73 st = 3.73 s

23. Part D
•
•
•
The ball does not return to the same height so we can’t use v
1
= -v
2

We need to add v
x
to v
y
to solve for the final velocity of the
ball. They are perpendicular vectors, so use Pythagorean
theorem
hyp
2
= (73.4)
2
+ (-21)
2
hyp = 76.3 m/s
X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2
v
x = 73.4 m/sv
x
= 73.4 m/sv
2 = -21 m/sv
2
= -21 m/s
V
x
= 73.4 m/s
V
y
=

-21 m/s
?
hyp

24. Part D
•
•
•
•
•
Now solve for the angle the ball strikes the ground.

tan α = opp
adj

α = tan
-1
(-21/73.4)
α = -16
o

The ball’s final velocity is 76.3 m/s at 16
o
below the
horizontal.

25. Sample Problem
•
•
•
An object is launched from the ground into the air at
an angle above the horizon, towards a vertical brick
wall that is 25.0 m horizontally from the launch point.
If the object takes 1.40 s to collide with the wall and
strikes the wall 5 m above the ground. Determine:
a) The object’s velocity in the horizontal direction
when it strikes the wall.
b) the velocity of the object when it is launched if it
strikes the wall 5 m above ground level

26. Solution
Part A
•
•
•
•
You have enough data to solve for velocity in
the x-direction using v = d/t, since a = 0.
v = d/t
v = 25/1.4
v = 17.9 m/s



X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2

d = 25 md = 25 m d = 5 md = 5 m
t = 1.4 st = 1.4 s
v = ?v = ? t = 1.4 st = 1.4 s

27. Solution
Part B
•
•
•
•
•
You have enough data to solve for the initial
velocity in the y-direction using d = v
1
t +
0.5at
2

d = v
1
t + 0.5at
2

5 = v
1
(1.4) + 0.5(-9.81)(1.4)
2
-V
1
(1.4) = -9.61 - 5
V
1
= 10.4 m/s

X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2

d = 25 md = 25 m d = 5 md = 5 m
t = 1.4 st = 1.4 s
v = ?v = ? t = 1.4 st = 1.4 s

28. Solution
Part B
•You know the initial velocity in the x and y
direction. You can use the Pythagorean theorem
and trig to solve for the size and angle of the
initial velocity.




X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2

d = 25 md = 25 m d = 5 md = 5 m
t = 1.4 st = 1.4 s
v = 17.9 m/sv = 17.9 m/st = 1.4 st = 1.4 s
?
hyp = ?
Opp
V
y
= 10.4 m/s
Adj
v
x
= 17.9 m/s

29. Solution
•
•
•
Solving for the size of the initial velocity:
hyp
2
= (10.4)
2
+ (17.9)
2
hyp = 20.7 m/s
Solving for the angle:
tan α = opp
adj
α = tan
-1
(10.4/17.9)
α = 30.2
o
The initial velocity is 20.7 m/s @ 30.2
o
above horizontal.
?
hyp = ?
Opp
V
y
= 10.4 m/s
Adj
v
x
= 17.9 m/s

30. Sample Problem
•
•
•
•
An object is launched from the ground into the air at
an angle of 28.0
o
(above the horizon) towards a
vertical brick wall that is 25.0 m horizontally from the
launch point. If the object takes 1.00 s to collide with
the wall, neglecting air resistance, determine:
a) The object’s velocity in the horizontal
direction when it strikes the wall.
b) The velocity of the object when it is
launched (remember velocity is a vector!)
c) The height at which the object strikes the
wall

31. Solution
Part A
•
•
•
•
You have enough data to solve for velocity in
the x-direction using v = d/t, since a = 0.
v = d/t
v = 25/1.00
v = 25.0 m/s



X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2

d = 25 md = 25 m
t = 1.00 st = 1.00 s
v = ?v = ? t = 1.00 st = 1.00 s

32. Solution
Part B
•You have been given the angle and one side, so
you can use trig. to solve for the hypotenuse,
which is the initial velocity.


X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2

d = 25 md = 25 m
t = 1.00 st = 1.00 s
v = 25 m/sv = 25 m/s t = 1.00 st = 1.00 s
28
o
hyp = ?
Opp
V
y
= ?
Adj
v
x
= 25 m/s

33. Solution
Part B
•
•
Solving for the size of the initial velocity:
cos α = adj
hyp
cos28 = 25
hyp
hyp = 28.3 m/s

The initial velocity is 28.3 m/s @ 28
o
above the
horizontal.
28
o
hyp = ?
Opp
V
y
= ?
Adj
v
x
= 25 m/s

34. Solution
Part C
•We can also use trig. to solve initial velocity in the y-
direction, which is the opposite side.
tan α = opp
adj
tan 28 = opp
25
The initial velocity in the y-direction is 13.3 m/s.
28
o
hyp = ?
Opp
V
y
= ?
Adj
v
x
= 25 m/s

35. Solution
Part C
•
•
•
•
You can now use position law #1 to
determine the height the object strikes the
wall.
d = v
1
t + 0.5at
2

d = 13.3(1.0) + 0.5(-9.81)(1.0)
2
d = 8.40 m

X-directionX-direction Y-directionY-direction
a = 0 m/s
2
a = 0 m/s
2
a = -9.81 m/s
2
a = -9.81 m/s
2

d = 25 md = 25 m V
1 = 13.3 m/sV
1
= 13.3 m/s
t = 1.00 st = 1.00 s
v = 25 m/sv = 25 m/s t = 1.00 st = 1.00 s

36. Sample Problem
•
•
Millen jumps off a 60.0 m high cliff. With what
velocity should he leave the cliff, (assume he jumps
horizontally) in order to just miss 12.0 m of rock
coming out from the cliff’s base? What is his final
velocity?
A = 3.43 m/s horizontally, 34.5 m/s 84.3
o
below
horizontal