What is projectile motion? Download

Transcript
What is projectile motion?
The only force acting on the objects above is the force of the
Earth.
With no Force of the Earth….
What do we actually see?
What do you notice about the horizontal and
vertical components of the motion?
What if we launch at an angle rather
than horizontally?
Suppose a rescue airplane
drops a relief package
while it is moving with a
constant horizontal speed
at an elevated height.
Assuming that air
resistance is negligible,
where will the relief
package land relative to
the plane?
a. below the plane and
behind it.
b. directly below the plane
c. below the plane and
ahead of it
What do you notice about the values
of Vx and Vy?
What do you notice about Vx and Vy
when we launch at an angle?
Kinematics Equations for Projectile
Motion:
• Y = ½ gt2 (This is vertical displacement of the
projectile when starting from rest)
• X = v0xt (This is horizontal displacement of the
projectile)
• Vfx = V0x Why is this true?
• Vfy = V0y + ayt
• Y = v0yt + ½gt2 if there is an initial velocity in the
y direction.
How do we solve these problems?
• The vertical motion of a projectile is just like
that of an object dropped or thrown straight
up or straight down. The force of the Earth is
acting on the object, so a = g (-9.8 m/s²).
• The horizontal motion of a projectile is just
like any constant velocity problem. No
horizontal forces are acting on the object so
Vₒₓ is the only velocity and aₓ = 0.
Now you try…
• Fill in the table below indicating the value of
the horizontal and vertical components of
velocity and acceleration for a projectile.
The answer:
Time (s)
Vx (m/s)
Vy (m/s)
Ax (m/s/s)
Ay (m/s/s)
0
1
2
3
15.0
15.0
15.0
15.0
29.4
19.6
9.8
0
0
0
0
0
-9.8
-9.8
-9.8
-9.8
4
5
6
15.0
15.0
15.0
-9.8
-19.6
-29.4
0
0
0
-9.8
-9.8
-9.8
The diagram to the right
shows the trajectory for a
projectile launched nonhorizontally from an
elevated position on top of
a cliff. The initial horizontal
and vertical components of
the velocity are 8 m/s and
19.6 m/s respectively.
Positions of the object at
1-second intervals are
shown. Determine the
horizontal and vertical
velocities at each instant
shown in the diagram.
The vx values will remain 8 m/s for the
entire 6 seconds.
The vy values will be changing by 9.8
m/s each second. Thus,
vy =9.8 m/s (t = 1 s) vy = 0 m/s (t = 2 s)
vy = -9.8 m/s (t = 3 s)
vy = -19.6 m/s (t = 4 s)
vy = -29.4 m/s (t = 5 s)
vy = -39.2 m/s (t = 6 s)
How do we find the initial component
velocities if we are given the angular
velocity?
Vₓ0 = V cos 60° = Vx
Vy0= V sin 60°
Notice that Vx2 + Vy2
= V2
What is the maximum height the ball
reaches?
• Ball reaches maximum height when Vy = 0, so
Vy = Vy0 + ayt becomes 0 = 43 m/s +(-9.8)t
-43 = -9.8t
t = 4.4 s; Y = ½ayt2 so Y = 94.3m
• Ball’s total hang time is 2 x the time it takes to
reach max height, so total time = 8.8 s.
• How far does the ball travel horizontally in the
time it is in the air?
Remember motion in the x direction is
independent of motion in the y
direction.
• X = X0 + Vx0t + ½ axt2
• X0 = 0 and ax = 0, so X = Vx0t = (25 m/s)(8.8s)=
220m.
So the projectile travels up and down with a max
height of 94.3 meters. When it lands back on the
ground, it is 220 m from its starting position.
Practice Problem:
• You accidentally throw your car
keys horizontally at 8.0 m/s from
a cliff 64-m high. How far from
the base of the cliff should you
look for your keys?
The answer:
•
•
•
•
V0x = 8.0m/s2 ,ax = 0, V0y = 0, ay = -9.8 m/s2
Yf = -64 m
- 64m = ½ (-9.8 m/s2)t2, t = 3.6 s
In that 3.6 seconds, how far do the keys travel
horizontally?
• X = V0xt=8.0m/s2(3.6s) = 28.8m
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