Vectors Projectiles
Friction on a Hope
Scalar -
magnitude (size) Reaction
90 to
·
Friction
An object
going under projectile motio
and horizontal component
T
magnitude& direction
Vector >
- ramp
·
Brick is in equilibrium
↓
displacement
Forces
) weight Mass of brick = 1
915kg Drag
distance
.
- -
A B A-B-C-D =
0
-
go
↓
=
I
A-D =
displacement down
- D
D . 92 96'R .
gW18 . 9N
Josie
S ⑤ =
18 67
& Sophie's force
.
# F2 96 .
if sophie &
A
Hawrai
Josies resultant
... F =
sin (9) x 18 9.
Force
force 2 96
naura puch H =
Terminal
.
josie with an
velocity ->
as
drag force ↑ acce
y max
EQUAL fore what velocity
object can reach
l velocity ↑
direction will the Mechanics
go
S displacement force
in ,
m vi = u2 + Las - once
drag =
gravitational po
V initial ms"
velocity Yterminal
v
a =
recultant force SCHTA velocity reached
hypotenuse > V final
velocity ms
-
-
A accelaration ms 2
direction of fore-Angle
s =
u+ + ka+ Mechanics and
Trig -
Ttime s
S = V+
Horizontal /vertical forces are EQUAL ↳
( =
mu
Materials
RequiredTractical
Y if constant acceleration 1 Set up the apparatus by attaching the e
Resolving Vectors Dacceleration of a tall clamp stand. Do not switch on th
,
v
J
② constant velocity area displacement
a is set up
2 Place the glass tube directly underneath
③ deceleration leaving space for the ball-bearing. Make
① constant flow t
3
Dconstant velocity downwards and not at an angle
S
velocity I
2
@ stationary 3 Attach both light gates around the glass
Fy SinOXT distance of around 10 cm
=
③ fast constant 4 Measure this distance between the two
&
velocity h with a metre ruler
5 Place the cushion directly underneath th
Ex =
COSOXF to catch the ball-bearing when it falls thr
6 Switch the current on the electromagnet
%90
x
30 T
Y 7
bearing directly underneath so it is attra
Turn the current to the electromagnet of
T
Friction on a Hope
Scalar -
magnitude (size) Reaction
90 to
·
Friction
An object
going under projectile motio
and horizontal component
T
magnitude& direction
Vector >
- ramp
·
Brick is in equilibrium
↓
displacement
Forces
) weight Mass of brick = 1
915kg Drag
distance
.
- -
A B A-B-C-D =
0
-
go
↓
=
I
A-D =
displacement down
- D
D . 92 96'R .
gW18 . 9N
Josie
S ⑤ =
18 67
& Sophie's force
.
# F2 96 .
if sophie &
A
Hawrai
Josies resultant
... F =
sin (9) x 18 9.
Force
force 2 96
naura puch H =
Terminal
.
josie with an
velocity ->
as
drag force ↑ acce
y max
EQUAL fore what velocity
object can reach
l velocity ↑
direction will the Mechanics
go
S displacement force
in ,
m vi = u2 + Las - once
drag =
gravitational po
V initial ms"
velocity Yterminal
v
a =
recultant force SCHTA velocity reached
hypotenuse > V final
velocity ms
-
-
A accelaration ms 2
direction of fore-Angle
s =
u+ + ka+ Mechanics and
Trig -
Ttime s
S = V+
Horizontal /vertical forces are EQUAL ↳
( =
mu
Materials
RequiredTractical
Y if constant acceleration 1 Set up the apparatus by attaching the e
Resolving Vectors Dacceleration of a tall clamp stand. Do not switch on th
,
v
J
② constant velocity area displacement
a is set up
2 Place the glass tube directly underneath
③ deceleration leaving space for the ball-bearing. Make
① constant flow t
3
Dconstant velocity downwards and not at an angle
S
velocity I
2
@ stationary 3 Attach both light gates around the glass
Fy SinOXT distance of around 10 cm
=
③ fast constant 4 Measure this distance between the two
&
velocity h with a metre ruler
5 Place the cushion directly underneath th
Ex =
COSOXF to catch the ball-bearing when it falls thr
6 Switch the current on the electromagnet
%90
x
30 T
Y 7
bearing directly underneath so it is attra
Turn the current to the electromagnet of
T
