Projectile Motion
Physics 6A
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Projectile motion is a combination of Horizontal and Vertical motion
We use separate sets of formulas for each,
but both motions happen simultaneously.
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Projectile motion is a combination of Horizontal and Vertical motion
We use separate sets of formulas for each,
but both motions happen simultaneously.
Horizontal
Vertical
Recall the formulas for linear motion
with constant acceleration:
(1)
x  x 0  v 0 t  21 at 2
(2)
v  v 0  at
(3)
v 2  v 02  2ax  x 0 
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Projectile motion is a combination of Horizontal and Vertical motion
We use separate sets of formulas for each,
but both motions happen simultaneously.
Horizontal
Vertical
Recall the formulas for linear motion
with constant acceleration:
In the case of projectiles, with no air resistance,
The horizontal motion is very simple:
The acceleration is zero, so the horizontal component
of the velocity is constant
(1)
x  x 0  v 0 t  21 at 2
(2)
v  v 0  at
(3)
v 2  v 02  2ax  x 0 
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Projectile motion is a combination of Horizontal and Vertical motion
We use separate sets of formulas for each,
but both motions happen simultaneously.
Horizontal
Vertical
In the case of projectiles, with no air resistance,
a = 0, so the formulas become:
x  x 0  v 0x t
v x  v 0x
ax  0
The horizontal component of V is constant!
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Projectile motion is a combination of Horizontal and Vertical motion
We use separate sets of formulas for each,
but both motions happen simultaneously.
Horizontal
Vertical
In the case of projectiles, with no air resistance,
The vertical component of the motion is just free-fall.
a = 0, so the formulas become:
This means the acceleration is constant, towards the
ground, with magnitude g = 9.8m/s2
x  x 0  v 0x t
v x  v 0x
ax  0
The horizontal component of V is constant!
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Projectile motion is a combination of Horizontal and Vertical motion
We use separate sets of formulas for each,
but both motions happen simultaneously.
Horizontal
Vertical
In the case of projectiles, with no air resistance,
The vertical component of the motion is just free-fall.
a = 0, so the formulas become:
This means the acceleration is constant, towards the
ground, with magnitude g = 9.8m/s2
x  x 0  v 0x t
v x  v 0x
ax  0
The horizontal component of V is constant!
Here are the formulas:
(1)
y  y0  v0y t  1 gt2
2
(2)
v y  v0y  gt
(3)
v2y  v20y  2gy  y0 
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Let’s try a couple of sample problems:
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Sample Problem #1
A baseball is thrown with an initial speed of 30 m/s, at an angle of 20° above the horizontal. When it leaves the thrower’s
hand the ball is 2.1 meters above the ground. In this problem we will ignore air resistance. Another player wants to
catch the ball, also at a height of 2.1 meters above ground. Where should he stand?
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Sample Problem #1
A baseball is thrown with an initial speed of 30 m/s, at an angle of 20° above the horizontal. When it leaves the thrower’s
hand the ball is 2.1 meters above the ground. In this problem we will ignore air resistance. Another player wants to
catch the ball, also at a height of 2.1 meters above ground. Where should he stand?
First we need to set up a coordinate system. A
convenient way to do it is to let the lowest point be
y=0, then call the upward direction positive.
With this choice, our initial values are:
y0 = 2.1m
v0 = 30m/s
V0 = 30 m/s
20°
y=2.1m
y=0
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Sample Problem #1
A baseball is thrown with an initial speed of 30 m/s, at an angle of 20° above the horizontal. When it leaves the thrower’s
hand the ball is 2.1 meters above the ground. In this problem we will ignore air resistance. Another player wants to
catch the ball, also at a height of 2.1 meters above ground. Where should he stand?
Since this is motion in 2 dimensions, we will want to find the
horizontal and vertical components of the initial velocity
v0x =
v0y =
V0 = 30 m/s
20°
y0=2.1m
y=0
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Sample Problem #1
A baseball is thrown with an initial speed of 30 m/s, at an angle of 20° above the horizontal. When it leaves the thrower’s
hand the ball is 2.1 meters above the ground. In this problem we will ignore air resistance. Another player wants to
catch the ball, also at a height of 2.1 meters above ground. Where should he stand?
Since this is motion in 2 dimensions, we will want to find the
horizontal and vertical components of the initial velocity
v0x = (30m/s)cos(20°) ≈ 28.2m/s
v0y = (30m/s)sin(20°) ≈ 10.3m/s
V0 = 30 m/s
20°
y0=2.1m
y=0
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Sample Problem #1
A baseball is thrown with an initial speed of 30 m/s, at an angle of 20° above the horizontal. When it leaves the thrower’s
hand the ball is 2.1 meters above the ground. In this problem we will ignore air resistance. Another player wants to
catch the ball, also at a height of 2.1 meters above ground. Where should he stand?
Since this is motion in 2 dimensions, we will want to find the
horizontal and vertical components of the initial velocity
v0x = (30m/s)cos(20°) ≈ 28.2m/s
v0y = (30m/s)sin(20°) ≈ 10.3m/s
V0,y ≈ 10.3m/s
V0,x ≈ 28.2m/s
y0=2.1m
y=0
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Sample Problem #1
A baseball is thrown with an initial speed of 30 m/s, at an angle of 20° above the horizontal. When it leaves the thrower’s
hand the ball is 2.1 meters above the ground. In this problem we will ignore air resistance. Another player wants to
catch the ball, also at a height of 2.1 meters above ground. Where should he stand?
Now we need to figure out how far the ball will travel horizontally. The only
relevant formula we have for horizontal motion is
x  x0  v0xt
We can use x0 = 0, and we just found v0,x
But what should we use for t?
V0,y ≈ 10.3m/s
V0,x ≈ 28.2m/s
y0=2.1m
y=0
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Sample Problem #1
A baseball is thrown with an initial speed of 30 m/s, at an angle of 20° above the horizontal. When it leaves the thrower’s
hand the ball is 2.1 meters above the ground. In this problem we will ignore air resistance. Another player wants to
catch the ball, also at a height of 2.1 meters above ground. Where should he stand?
The time will be the same as the time it takes to travel up and then back
down to a height of 2.1m. We need to use the vertical motion to find this.
V0,y ≈ 10.3m/s
V0,x ≈ 28.2m/s
y0=2.1m
y=0
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Sample Problem #1
A baseball is thrown with an initial speed of 30 m/s, at an angle of 20° above the horizontal. When it leaves the thrower’s
hand the ball is 2.1 meters above the ground. In this problem we will ignore air resistance. Another player wants to
catch the ball, also at a height of 2.1 meters above ground. Where should he stand?
The time will be the same as the time it takes to travel up and then back
down to a height of 2.1m. We need to use the vertical motion to find this.
There are a few options on how to proceed. In this case a simple and
direct way is just to use the basic vertical position equation:
y  y0  v0y t  1 gt2
2
V0,y ≈ 10.3m/s
V0,x ≈ 28.2m/s
y0=2.1m
y=0
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Sample Problem #1
A baseball is thrown with an initial speed of 30 m/s, at an angle of 20° above the horizontal. When it leaves the thrower’s
hand the ball is 2.1 meters above the ground. In this problem we will ignore air resistance. Another player wants to
catch the ball, also at a height of 2.1 meters above ground. Where should he stand?
Here is the calculation:
y  y0  v 0,y t  1 gt2
2
 
 4.9 m   t 2  10.3 m  t  0


s
s 


m   t  10.3 m   t  0
4
.
9



s 
s 
2.1m  2.1m  10.3 m  t  1  9.8 m2   t 2
s
2
s 
2
2
t  2.1s
V0,y ≈ 10.3m/s
V0,x ≈ 28.2m/s
y0=2.1m
y=0
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Sample Problem #1
A baseball is thrown with an initial speed of 30 m/s, at an angle of 20° above the horizontal. When it leaves the thrower’s
hand the ball is 2.1 meters above the ground. In this problem we will ignore air resistance. Another player wants to
catch the ball, also at a height of 2.1 meters above ground. Where should he stand?
Now that we have the time, we can use our horizontal equation:
x  x 0  v 0,x t
x  0  28.2 ms   2.1s 
x  59m
V0,y ≈ 10.3m/s
V0,x ≈ 28.2m/s
y0=2.1m
x
y=0
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Sample Problem #2
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball horizontally at 8.50m/s, and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
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Sample Problem #2
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball horizontally at 8.50m/s, and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
Here is a picture of the situation. We can set up our coordinate system with the origin at the base of the tree.
y
12m
V0
Vdog = ?
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Sample Problem #2
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball horizontally at 8.50m/s, and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
Here is a picture of the situation. We can set up our coordinate system with the origin at the base of the tree.
Part (a) of this problem is easy if we remember that the horizontal and vertical motions of the ball are independent.
There is no acceleration in the x-direction, so the ball will have a constant horizontal component of velocity. If the
dog is going to catch the ball, his horizontal velocity must be the same as the ball (the ball will be directly above
the dog the whole time). So our answer is 8.50 m/s.
y
12m
V0
Vdog = V0
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Sample Problem #2
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball horizontally at 8.50m/s, and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
For part (b) we need to use our free-fall formulas to find out how long it takes for the ball to get to the ground.
Then we can use the horizontal equation to find the distance:
(1)
y  y0  v0y t  1 gt2
2
(2)
v y  v0y  gt
(3)
v2y  v20y  2gy  y0 
Which of these equations
should we use?
y
12m
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Sample Problem #2
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball horizontally at 8.50m/s, and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
For part (b) we need to use our free-fall formulas to find out how long it takes for the ball to get to the ground.
Then we can use the horizontal equation to find the distance:
(1)
y  y0  v0y t  1 gt2 Equation (1) works because we
2
(2)
v y  v0y  gt
(3)
v2y  v20y  2gy  y0 
know the initial and final height.
y
12m
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Sample Problem #2
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball horizontally at 8.50m/s, and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
For part (b) we need to use our free-fall formulas to find out how long it takes for the ball to get to the ground.
Then we can use the horizontal equation to find the distance:
 
0m  12m  0 m  t  1  9.8 m2 t 2
s
2
s 
t  1.56s
y
12m
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Sample Problem #2
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball horizontally at 8.50m/s, and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
For part (b) we need to use our free-fall formulas to find out how long it takes for the ball to get to the ground.
Then we can use the horizontal equation to find the distance:
 
0m  12m  0 m  t  1  9.8 m2 t 2
s
2
s 
t  1.56s
x  x0  v 0,x  t


x  0m  8.50 m  1.56s   13.3m
s
For an extra bonus challenge, try this problem
again but have the boy throw the ball at an angle
of 20° above the horizontal at speed 8.50 m/s.
This is worked out on the following slides, but
please try it on your own first.
y
12m
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Sample Problem #2 (bonus version)
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball at 8.50m/s, at an angle of 20°,and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
Here is a picture of the situation. We set up our coordinates with the origin at the base of the tree.
y
12m
Vdog = ?
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Sample Problem #2 (bonus version)
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball at 8.50m/s, at an angle of 20°,and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
Here is a picture of the situation. We set up our coordinates with the origin at the base of the tree.
The first thing we have to do is break the initial velocity into components:
v0x = (8.50m/s)cos(20°) ≈ 7.99m/s
v0y = (8.50m/s)sin(20°) ≈ 2.91m/s
y
12m
Vdog = ?
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Sample Problem #2 (bonus version)
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball at 8.50m/s, at an angle of 20°,and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
Here is a picture of the situation. We set up our coordinates with the origin at the base of the tree.
The first thing we have to do is break the initial velocity into components:
v0x = (8.50m/s)cos(20°) ≈ 7.99m/s
v0y = (8.50m/s)sin(20°) ≈ 2.91m/s
Now part a) is just like before: The dog must run at 7.99 m/s to keep up with the ball.
y
12m
Vdog = ?
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Sample Problem #2 (bonus version)
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball at 8.50m/s, at an angle of 20°,and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
Here is a picture of the situation. We set up our coordinates with the origin at the base of the tree.
The first thing we have to do is break the initial velocity into components:
v0x = (8.50m/s)cos(20°) ≈ 7.99m/s
v0y = (8.50m/s)sin(20°) ≈ 2.91m/s
Now part a) is just like before: The dog must run at 7.99 m/s to keep up with the ball.
For part b) we can use our y-position formula again, although this time we might need the
quadratic equation to solve it:
y


0m  12m  2.91 m  t  1  9.8 m2 t2
s
2
s 
t  1.89s
12m
Vdog = ?
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Sample Problem #2 (bonus version)
A boy 12.0m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running at a constant
speed the instant the ball is thrown. The boy throws the ball at 8.50m/s, at an angle of 20°,and air resistance can be ignored.
(a) How fast must the dog run to catch the ball just as it reaches the ground?
(b) how far from the tree will the dog be when it catches the ball?
Here is a picture of the situation. We set up our coordinates with the origin at the base of the tree.
The first thing we have to do is break the initial velocity into components:
v0x = (8.50m/s)cos(20°) ≈ 7.99m/s
v0y = (8.50m/s)sin(20°) ≈ 2.91m/s
Now part a) is just like before: The dog must run at 7.99 m/s to keep up with the ball.
For part b) we can use our y-position formula again, although this time we might need the
quadratic equation to solve it:
y


0m  12m  2.91 m
t 
s
1
2
9.8 t
m
s2
2
t  1.89s
x  x0  v0,x  t
12m


x  0m  7.99 m
 1.89s  15.1m
s
So the ball lands 15.1 meters from the tree.
Vdog = ?
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