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A Projectile Fired at an Angle
A conceptual study
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When an object is dropped, the only thing that is affecting its motion is gravity, g.
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When an object is dropped, the only thing that is affecting its motion is gravity, g.
Gravity is a force, F.
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When an object is dropped, the only thing that is affecting its motion is gravity, g.
Gravity is a force, F.
Newton says that a force causes a mass, m, to accelerate:
maF
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So, when an object is dropped, it starts from rest (vi = 0 m/s), and it accelerates at ~+10 m/s each second.
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So, when an object is dropped, it starts from rest (vi = 0 m/s), and it accelerates at ~+10 m/s each second.
So, each second, the object is traveling 10 m/s faster than it was the second before.
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Also, since the object is accelerating, it will travel a greater distance each second than it did the second before.
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Also, since the object is accelerating, it will travel a greater distance each second than it did the second before.
So, it’s obvious that, the longer an object falls, the faster it will go and the farther it will fall.
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Let’s consider an object that’s thrown straight up.
It goes up and then slows to a stop.
At its highest point, its instantaneous velocity is 0 m/s.
It then starts downward as if dropped from rest at that height.
0 m/s
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During the upward part of the motion, the object slows from its initial upward velocity, vi, to 0 velocity.
0 m/s
-g +g
-vi +vi
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During the upward part of the motion, the object slows from its initial upward velocity, vi, to 0 velocity.
0 m/s
-g +g
-vi +vi
We know its accelerating because its velocity is changing.
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During the upward part of the motion, the object slows from its initial upward velocity, vi, to 0 velocity.
0 m/s
-g +g
-vi +vi
We know its accelerating because its velocity is changing.
Since it’s decreasing in velocity, we say it’s accelerating negatively.
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Each second, it is moving at 10 m/s slower than it was the second before.
0 m/s
-40 m/s
-30 m/s
-20 m/s
-10 m/s
40 m/s
30 m/s
20 m/s
10 m/s
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Each second, it is moving at 10 m/s slower than it was the second before.
0 m/s
-40 m/s
-30 m/s
-20 m/s
-10 m/s
40 m/s
30 m/s
20 m/s
10 m/s
Once it reaches the top and begins to fall back, it is moving at 10 m/s faster each second.
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Each second, it is moving at 10 m/s slower than it was the second before.
0 m/s
-40 m/s
-30 m/s
-20 m/s
-10 m/s
40 m/s
30 m/s
20 m/s
10 m/s
Once it reaches the top and begins to fall back, it is moving at 10 m/s faster each second.
Whether moving up or down, the object accelerates at the same rate…10 m/s2.
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Now, let’s consider objects that are thrown up and horizontally at the same time, a projectile fired at an angle.
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Now, let’s consider objects that are thrown up and horizontally at the same time, a projectile fired at an angle. The objects’
vertical and horizontal motions are completely independent of each other.
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The object will move vertically exactly the same as it would if thrown straight up.
Vertical motion
Horizontal motion
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The object will move vertically exactly the same as it would if thrown straight up.
It will move horizontally just like it would if it were rolling across a smooth, level, surface.
Vertical motion
Horizontal motion
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It will accelerate vertically.It will move at a constant velocity horizontally.g
v
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Let’s say Kyle threw a ball at some velocity, v, at 60° above the horizontal.
60°
v
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Let’s say Kyle threw a ball at some velocity, v, at 60° above the horizontal.
60°
v
The thrown ball is moving upward and horizontally at the same time.
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Let’s say Kyle threw a ball at some velocity, v, at 60° above the horizontal.
60°
v
The thrown ball is moving upward and horizontally at the same time. The ball has
an initial vertical velocity, vy.
vy
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Let’s say Kyle threw a ball at some velocity, v, at 60° above the horizontal.
60°
v
The thrown ball is moving upward and horizontally at the same time. The ball has
an initial vertical velocity, vy.
vy It also has an initial horizontal velocity, vx.
vx
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vy = vsin
60°
vvy
vx
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vy = vsin
60°
vvy
vx
vx = vcos
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vy = vsin
60°
v = 50 m/svy
vx
vx = vcos
Let’s say Kyle through the ball at 50 m/s. Not bad!
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vy = vsin
60°
v = 50 m/svy
vx
vx = vcos
Let’s say Kyle through the ball at 50 m/s. Not bad!
So what’s the horizontal and vertical velocities of the throw?
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60°
v = 50 m/svy = 43 m/s
vx
m/s 43
)60sin m/s)( 50(
sin
y
y
y
v
v
vv
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60°
v = 50 m/svy = 43 m/s
vx = 25 m/s
m/s 43
)60sin m/s)( 50(
sin
y
y
y
v
v
vv
m/s 25
)60 cos m/s)( 50(
cos
x
x
x
v
v
vv
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60°
v = 50 m/svy = 43 m/s
vx = 25 m/s
What should happen to the horizontal velocity as time passes?
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60°
v = 50 m/svy = 43 m/s
vx = 25 m/s
What should happen to the horizontal velocity, vx, as time passes?
Nothing!
There is no horizontal force, therefore, no horizontal acceleration.
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vx
vx
vx
vx
vx
vx
vx
vx
vx
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60°
v = 50 m/svy = 43 m/s
vx = 25 m/s
What should happen to the vertical velocity, vy, as time passes?
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60°
v = 50 m/svy = 43 m/s
vx = 25 m/s
What should happen to the vertical velocity, vy, as time passes?
It should decrease until it reaches the top of its flight, then increase as it falls.
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vy
vy
vy
vy
vy
vy
vy
vy
vy = 0
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60°
v = 50 m/svy = 43 m/s
vx = 25 m/s
What should happen to the vertical acceleration as time passes?
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60°
v = 50 m/svy = 43 m/s
vx = 25 m/s
What should happen to the vertical acceleration as time passes?
Nothing. Acceleration is gravity, g, and remains at 10 m/s2.
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g
gg
g
g
g = 10 m/s
g
g
g