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•  Tutorial homework is due at the beginning of your recitation on Thursday/Friday.

•  The first midterm is two weeks from tomorrow.

Web page: http://www.colorado.edu/physics/phys1110/

Announcements: Projectile motion

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Clicker question 1 Q. A particle is moving at constant speed along the path shown. Its velocity vector at two different times is shown. What is the direction of the acceleration when the particle is at point X?

Set frequency to BA

D

C

B

A

(E) None of these

X

tvaΔΔ=!! Since it points in the same

direction as =Δv! − =

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Also, y = y0 +12 vy0 + vy( )Δt and vy

2 = vy02 − 2g Δy

so x = x0 + vx0Δt and vx = vx0

Projectile motion This describes the motion of a body (bullet, basketball, motorcycle, etc.) in free fall after being launched.

The only acceleration is due to gravity and is always straight down.

Thus, the velocity in the horizontal direction is constant

In the vertical direction there is acceleration from gravity

so with ay = −g then y = y0 + vy0Δt −12 g Δt( )2 and vy = vy0 − g Δt

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Clicker question 2 Q. A basketball is launched from a basketball cannon and follows the trajectory shown. What is the direction of the acceleration at point X?

Set frequency to BA

D

C

BA

X

E

Once launched, the only acceleration is due to gravity and is straight down.

The horizontal velocity is constant

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Initial velocity in horizontal and vertical directions depends on angle θ0 and speed v0 of launch.

Velocities in projectile motion

θ0 vx0 = v0 cosθ0 vy0 = v0 sinθ0and

Solving projectile motion problems

Realize the horizontal and vertical motions are independent. Their only connection is through the time the projectile is in the air.

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Clicker question 3 Set frequency to BA Q. Two balls are thrown at the same time with the same speed (10 m/s). One ball is thrown straight up and the other ball is thrown at an angle of 45°. Which ball hits the ground first? A.  The ball thrown straight up B.  The ball thrown at an angle C.  Both hit at the same time D.  Not enough information to tell

y = y0 + vy0Δt −12 g Δt( )2

Solve with

y = y0 = 0 0 = 0+ vy0Δt −12 g Δt( )2 vy0 =

12 g Δt Δt =

2vy0gso so so

The ball thrown straight up: vy0=10 m/s so Δt = 2 ⋅10 m/s9.8 m/s2 = 2.04 s

The ball thrown at an angle has Δt = 2 ⋅ 7.07 m/s9.8 m/s2 =1.44 svy0 = v0 sinθ0 =10 m/s ⋅sin 45° = 7.07 m/s

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Analyzing the vertical motion In the previous problem we had an initial upward velocity vy0 and asked how long it took to come back to Earth.

Using with y=y0 we found: y = y0 + vy0Δt −12 g Δt( )2 Δt =

2vy0g

How long does it take to reach the top of its trajectory?

vy = vy0 − g ΔtUsing with vy=0 we find: Δt =vy0g

So the projectile spends half the time going up and half the time going down. (This is only true if y=y0.)

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Solving a projectile motion problem

θ0

A basketball launched on a level surface travels 15.0 m and reaches a maximum height of 6.40 m. What is the initial velocity ?

!v0 = v0,θ0( )1. Draw a diagram

2. Figure out what we know

3. Figure out what we need:

Need vx0 & vy0 to get v0 & θ0

0v!

00 =x

0

1

2

00 =y

ay0 = −9.8 m/s2

vx0 = ?vy0 = ?

00 =t

m 152 =x02 =y

ay2 = −9.8 m/s2

vx2 = vx0vy2 = ?

?2 =t

?1 =xm 4.61 =y

ay1 = −9.8 m/s2

vx1 = vx0vy1 = 0

?1 =t

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Solving a projectile motion problem Basketball launched on a level surface travels 15.0 m and reaches a maximum height of 6.40 m. What is the initial velocity ?

!v0 = v0,θ0( )

To get v0x we first need the flight time to be able to use x = x0 + vx0Δt

Can get v0y from with y=y1, vy=0, vy0=vy0 vy2 = vy0

2 − 2g y− y0( )vy0 = 2g Δy = 2(9.8 m/s2 )(6.4 m-0 m) =11.2 m/s

Can now use y = y0 + vy0Δt −12 g Δt( )2 to get Δt

y2 = y0 + vy0Δt02 −12 g Δt02( )2 Δt =

2vy0

g=

2 ⋅11.2 m/s9.8 m/s2 = 2.29 swith y=y0:

θ0 0

1

2

vx0 =ΔxΔt

=15 m2.29 s

= 6.56 m/sNow we can get vx0:

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Answer to projectile motion problem Projectile follows path shown reaching a maximum height of 6.40 m and distance of 15.0 m. Find the initial velocity:

!v0 = v0,θ0( )

Therefore v0 = vx0

2 + vy02 = (6.56 m/s)2 + (11.2 m/s)2 =13.0 m/s

θ0 = tan−1 11.2 m/s6.56 m/s"

#$

%

&'= 59.6°

vy0 =11.2 m/svx0 = 6.56 m/s

We found that

Can think of this as a ball moving at a constant 6.56 m/s in the x direction. The ball stops when it hits the ground. This is determined by considering how long a ball thrown up at 11.2 m/s stays in the air.

θ0 0

1

2

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Some projectile motion warnings

•  The time of flight is twice the time to reach the maximum height.

•  The initial and final vertical velocities have the same magnitude and opposite directions.

•  Maximum range is obtained with an angle of 45°

Following are true only when !!!!! 0=Δy(when the launch and landing points are at the same height)

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Clicker question 4 Set frequency to BA

Q. A rifle is accurately aimed at a rabid monkey hanging from the branch of a tree. The instant the gun is fired, the monkey releases the branch and starts falling. The monkey is well within the range of the rifle. The initial speed of the bullet is vo. What happens?

v0 gun

A: The bullet finds its target, regardless of the value of vo. (Assuming vo is large enough to reach the air below the monkey.)

B: The bullet hits the monkey only if vo is very large so the monkey does not fall very far.

C: The bullet misses.