t = 2H / g x = v0t = v0 2H / g I. During its entire flight, the minimum speed of the bullet is v0.

II. The acceleration of the bullet is constant during its flight.

III. The time it takes for the bullet to hit the ground increases as v0 increases.

A) All of the above statements are true.

B) Only two of the statements are true.

C) Only one statement is true.

D) None is true.

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I & II

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Assignments



For this week:

•  Read Ch. 4 of Wolfson and Prof Dubson’s notes.

•  Do Recitation Homework #3 (found on course web site under Tutorials, Assignments) by your Wed/Thurs recitation.

•  CAPA 4 is posted (problems on Newton’s laws).

•  Prof Munsat recommends the following web site for students who’d like more review of vectors: http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html

Exam next Thursday :

•  Locations presented in class next Monday.

•  Old exam (and solns) on D2L.

•  Begin to prepare now! See course web site…..

Last time:

•  Derived and began to apply the kinematic equations in 2D.

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Participate in Research on Learning and Understanding Physics!

As part of a study on the ideas and understanding that physics students have about energy, we are looking for CU students who are willing to discuss several questions about energy. This research investigates the productive ideas that students have, with the intent of providing information that will support instructors in effectively presenting and teaching the concept of energy.

Your participation would involve one interview, in which you would answer several questions about energy. Interviews are expected to last about 45 minutes.

While you will not receive any direct benefit from participation, you will help the physics community to better understand your thinking and help us gather information that may help instructors tailor their instruction to students’ needs. Additionally, the interviewers are willing to answer any physics questions you may have or discuss their experience as physics majors with you.

If interested, please contact Lisa Goodhew: [email protected] *free donuts at interview sessions!

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Kinematic Equations in 2D

v0

+y θ

y0

vy0

v x0

Kinematic equations for projectiles subject to gravity without air resistance.

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Vertical y-direction

ay = −g y0 v0 y = v0 sin θ

Horizontal x-direction

ax = 0 x0 = 0 v0 x = v0 cosθ

vy (t) = v0 sin θ − gt

vx (t) = v0 cosθ

1 2 y(t) = y0 + v0 yt − gt 2 2 vy2 (y) = vy0 − 2gΔy

x(t) = v0 x t vx2 (x) = vx20

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Summary: Ground to Ground Projectile

H

0 = x0 = y0

R

You derived three formulas about the behavior of a projectile moving from ground level to ground level:

t max

v0 sin θ =2 =2 g g vy0

v02 sin 2 θ H= = 2g 2g 2vx 0 vy0 v02 sin 2θ R= = g g 2 vy0

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Time of flight

Height of apex

“Range” – horizontal distance traveled.

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(1) t max

tmax=2.1s

2v sin θ = 0 g

v02 sin 2 θ (2) H = 2g

H=5.5 m

v02 sin 2θ (3) R = g

 v0

R = 270 ft = 83 m Travis Pastrana, 2009

v0 = 90mi / h = 40m / s

θ=15°

Rg (83)(9.9)  (3) ⇒ sin 2θ = 2 = = 0.51 ⇒ θ = 15 v0 (40)2

v02 sin 2 θ (40)2 sin 2 15 (2) ⇒ H = = m = 5.5m 2g 19.6 2v0 sin θ 2(40)(sin15 ) (1) ⇒ t max = = s = 2.1s g 9.8 L10      W  9/16/14  

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(1) t max

2v sin θ = 0 g

v02 sin 2 θ (2) H = 2g

v02 sin 2θ (3) R = g v0 ~ 250mi / h~120m / s

 v sin 45 v0 0

tmax?

H? R = 1 mile = 1.6 km Kenny Powers, 1976



~ 85m / s

θ=45°

(120)2 (3) ⇒ R = m ~ 1440m 9.8 (85)2 (2) ⇒ H = m ~ 360m 19.6 2(85) (1) ⇒ t max = s ~ 17s 9.8

http://www.youtube.com/watch?v=GLsVWFGO7aQ L10      W  9/16/14  

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   r = r0 + v0t  v0t  r0

 r

   r = r0 + v0t Final position (half way to the monkey)

Initial position (hunter)

Change in position

 r Remember that the position vector is always drawn relative to the origin.

(D) None of the above.

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y

A gun is aimed right at a naughty monkey in a tree. The instant the gun is fired the monkey drops from the tree and begins to fall. The monkey is well within range of the gun. What happens?

H

 vo θ

x bullet & the monkey fall A) The bullet will hit the monkey.

The at the same rate and, B) The bullet will miss high.

therefore, the same distance. C) The bullet will miss low.

D) Impossible to tell with the information given.

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Monkey and the Hunter

y

 g = −gyˆ 1 2 gt 2

 v0t

 r  r0 = (0,0)

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1 2    r = r0 + v0t + at 2 1 2   r = v0t + gt 2

1 2 Bullet drops a distance =

gt 2 How far does the monkey drop x in the same amount of time?

1 2 gt 2

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y

Monkey and the Hunter

What’s the angle θ to hit the monkey?

ymonkey = ybullet at x = R.

H

 vo θ

x

R

1 ymonkey = H − gt 2 2 1 ybullet = v0 yt − gt 2 2 Time for the bullet to get to

t(R) = R / v0 x x=R.

ymonkey = ybullet

⇒

1 1 H − gt 2 = v0 yt − gt 2 2 2 ⎛ R⎞ ⎛ v0 sin θ ⎞ H = v0 yt = v0 y ⎜ ⎟ = R ⎜ = R tan θ ⎟ ⎝ v0 x ⎠ ⎝ v0 cosθ ⎠ H tan θ = R

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⇒

Aim right at the monkey!

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From Schuyler Nippert:

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From Schuyler Nippert:

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True or False.



If the speed of an object is constant its acceleration must be zero.



A) True







B) False

 Δv  a = lim Δt→0 Δt

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Acceleration depends on the change in the velocity vector.

Velocity vector can change in two ways.

 1)  v (speed) can change.

 2)  Direction of v can change.

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True or False.



If the speed of an object is constant its acceleration must be zero.



A) True







B) False

 Δv  a = lim Δt→0 Δt

Acceleration depends on the change in the velocity vector.



Velocity vector can change in two ways:

 1)  v (speed) can change.

 2)  Direction of v can change.

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 v2 2

(E) None of these.

 Δv

 v2

 v1 1

Therefore, acceleration is toward the center of the circle.

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    Δv v2 − v1 a≈ = Δt Δt    v1 + Δv = v2 17  

 Circular Motion with Constant Speed

v = v  v Velocity vector has constant length.

It points tangent to the circle.

 v



center of the circle.



x

 v

Acceleration vector points toward the

T = time or period of 1 rev

r

 a

v = distance/time

= circumference/period = 2πr/T

 a:    the center (centripetal accel) dir of a = dir of Δv = toward v  2 a = a = v /r L10      W  9/16/14  

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