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.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Projectile motion is a combination of Horizontal and Vertical motion

We use separate sets of formulas for each,

but both motions happen simultaneously.

Horizontal

Recall the formulas for linear motion

with constant acceleration:

Vertical

020

2

0

221

00

xxa2vv

atvv

attvxx

(1)

(2)

(3)

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For Campus Learning Assistance Services at UCSB

Projectile motion is a combination of Horizontal and Vertical motion

We use separate sets of formulas for each,

but both motions happen simultaneously.

Horizontal

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

Vertical

020

2

0

221

00

xxa2vv

atvv

attvxx

(1)

(2)

(3)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Projectile motion is a combination of Horizontal and Vertical motion

We use separate sets of formulas for each,

but both motions happen simultaneously.

Horizontal

In the case of projectiles, with no air resistance,

a = 0, so the formulas become:

Vertical

The horizontal component of V is constant!

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For Campus Learning Assistance Services at UCSB

0a

vv

tvxx

x

x0x

x00

Projectile motion is a combination of Horizontal and Vertical motion

We use separate sets of formulas for each,

but both motions happen simultaneously.

Horizontal

In the case of projectiles, with no air resistance,

a = 0, so the formulas become:

Vertical

The vertical component of the motion is just free-fall.

This means the acceleration is constant, towards the ground, with magnitude g = 9.8m/s2

The horizontal component of V is constant!

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

0a

vv

tvxx

x

x0x

x00

Projectile motion is a combination of Horizontal and Vertical motion

We use separate sets of formulas for each,

but both motions happen simultaneously.

Horizontal

In the case of projectiles, with no air resistance,

a = 0, so the formulas become:

Vertical

The vertical component of the motion is just free-fall.

This means the acceleration is constant, towards the ground, with magnitude g = 9.8m/s2

Here are the formulas:

The horizontal component of V is constant! 0

2y0

2y

y0y

221

y00

yyg2vv

gtvv

gttvyy

(1)

(2)

(3)

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0a

vv

tvxx

x

x0x

x00

Let’s try a couple of sample problems:

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Sample Problem #1A 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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y=0

y=2.1m

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

Sample Problem #1A 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?

V0 = 30 m/s

20°

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y=0

y0=2.1m

Since this is motion in 2 dimensions, we will want to find the horizontal and vertical components of the initial velocity

v0x =

v0y =

Sample Problem #1A 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?

V0 = 30 m/s

20°

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y=0

y0=2.1m

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

Sample Problem #1A 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?

V0 = 30 m/s

20°

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y=0

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

Sample Problem #1A 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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V0,x ≈ 28.2m/s

V0,y ≈ 10.3m/s

y0=2.1m

y=0

Sample Problem #1A 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?

V0,x ≈ 28.2m/s

V0,y ≈ 10.3m/s

Now we need to figure out how far the ball will travel horizontally. The only relevant formula we have for horizontal motion is

tvxx x00

We can use x0 = 0, and we just found v0,x

But what should we use for t?

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y0=2.1m

Sample Problem #1A 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.

y=0

V0,x ≈ 28.2m/s

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V0,y ≈ 10.3m/s

y0=2.1m

Sample Problem #1A 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?

V0,y ≈ 10.3m/s

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:

221

y00 gttvyy

y=0

V0,x ≈ 28.2m/s

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y0=2.1m

Sample Problem #1A 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?

V0,y ≈ 10.3m/s

Here is the calculation:

s

sm

sm

sm2

sm

2

sm

21

smmm

221

y,00

1.2t

0t3.10t9.4

0t3.10t9.4

t8.9t3.101.21.2

gttvyy

2

2

2

y=0

y0=2.1m

V0,x ≈ 28.2m/s

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For Campus Learning Assistance Services at UCSB

Sample Problem #1A 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?

V0,y ≈ 10.3m/s

Now that we have the time, we can use our horizontal equation:

y=0

y0=2.1m

V0,x ≈ 28.2m/s

m

ssm

x,00

59x

1.22.280x

tvxx

x

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Sample Problem #2A 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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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?

V0

Here is a picture of the situation. We can set up our coordinate system with the origin at the base of the tree.

x

y

Vdog = ?

12m

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Sample Problem #2

Vdog = V0

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.

x

y

12m 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:

x

y

12m

Which of these equations should we use?

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Sample Problem #2

02

y02y

y0y

221

y00

yyg2vv

gtvv

gttvyy

(1)

(2)

(3)

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:

x

y

12m

Equation (1) works because we know the initial and final height.

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For Campus Learning Assistance Services at UCSB

Sample Problem #2

02

y02y

y0y

221

y00

yyg2vv

gtvv

gttvyy

(1)

(2)

(3)

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:

x

y

12m

s56.1t

t8.9t0m12m0 2

sm

21

sm

2

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For Campus Learning Assistance Services at UCSB

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:

x

y

12m

m3.13s56.150.8m0x

tvxx

s56.1t

t8.9t0m12m0

sm

x,00

2

sm

21

sm

2

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.

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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 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?

V 0

Here is a picture of the situation. We set up our coordinates with the origin at the base of the tree.

x

y

Vdog = ?

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?

V 0

Here is a picture of the situation. We set up our coordinates with the origin at the base of the tree.

x

y

Vdog = ?

12m

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Sample Problem #2 (bonus version)

The first thing we have to do is break the initial velocity into components:v0x = (8.50m/s)cos(20°) ≈ 7.99m/sv0y = (8.50m/s)sin(20°) ≈ 2.91m/s

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?

V 0

Here is a picture of the situation. We set up our coordinates with the origin at the base of the tree.

x

y

Vdog = ?

12m

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For Campus Learning Assistance Services at UCSB

Sample Problem #2 (bonus version)

Now part a) is just like before: The dog must run at 7.99 m/s to keep up with the ball.

The first thing we have to do is break the initial velocity into components:v0x = (8.50m/s)cos(20°) ≈ 7.99m/sv0y = (8.50m/s)sin(20°) ≈ 2.91m/s

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?

V 0

Here is a picture of the situation. We set up our coordinates with the origin at the base of the tree.

x

y

Vdog = ?

12m

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For Campus Learning Assistance Services at UCSB

Sample Problem #2 (bonus version)

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:

s89.1t

t8.9t91.2m12m0 2

sm

21

sm

2

The first thing we have to do is break the initial velocity into components:v0x = (8.50m/s)cos(20°) ≈ 7.99m/sv0y = (8.50m/s)sin(20°) ≈ 2.91m/s

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?

V 0

Here is a picture of the situation. We set up our coordinates with the origin at the base of the tree.

x

y

Vdog = ?

12m

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For Campus Learning Assistance Services at UCSB

Sample Problem #2 (bonus version)

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:

m1.15s89.199.7m0x

tvxx

s89.1t

t8.9t91.2m12m0

sm

x,00

2

sm

21

sm

2

So the ball lands 15.1 meters from the tree.

The first thing we have to do is break the initial velocity into components:v0x = (8.50m/s)cos(20°) ≈ 7.99m/sv0y = (8.50m/s)sin(20°) ≈ 2.91m/s