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