Save
science projevtion
Save
Share
Learn
Content
Leaderboard
Learn
Created by
JAMES GULAC
Visit profile
Subdecks (1)
SCIENCE LESSON 2
science projevtion
12 cards
Cards (32)
Projectile motion
Combination of horizontal motion with
constant
velocity and vertical motion with
constant
acceleration
Horizontal
component of projectile motion
Acceleration equal to
zero
(ax=0) since the horizontal component of the velocity of projectiles is
constant
Vertical component
of
acceleration
Constant, equal to the
acceleration
due to
gravity
(ay= -g)
Equations
of motion with respect to x and
y-axis
x
= x0 + v0xt
y = y0 +
v0yt
-
1/2
gt^2
Equations
for velocities
vx
=
v0x
vy =
v0y - gt
Factors
affecting height and range of a projectile
Initial
velocity
(v0)
Angle
of launch (θ)
Equations
for x and y components of initial velocity
v0x =
v0 cos
θ
v0y
= v0
sin
θ
Equations of motion for final position
x =
x0
+
v0x
t
y =
y0
+
v0y
t - 1/2 gt^2
Conditions for projectile motion: Object thrown with
initial velocity v0
at angle of
launch θ
Trajectory of projectile
Parabolic
At peak of
parabola
, vertical component of velocity is
zero
(vy=0)
Time
when projectile reaches maximum height
t =
v0y/g
Maximum height (H)
H =
v0y
^2 /
2g
Time
to go down to same level (tdown) = time to go
up
(tup)
Total time of flight (T)
T =
2t
Equation
for maximum
range
(R)
R = (
v0
^
2
* sin(2θ)) / g
Range and
height
depend on angle of
launch
(θ)
Salient
values of angle θ
θ = 0°:
Horizontal
motion only, height = 0
θ = 90°:
Vertical
motion only,
range = 0
1° < θ < 89°:
Projectile
motion evident
When θ =
45°
,
horizontal
displacement is maximum with high vertical displacement
When θ =
89°
,
vertical
displacement is maximum with low horizontal displacement
See all 32 cards