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Physics 140 – Fall 2014September 15

Reminder:

• Mastering Physics homework #1 due tomorrow night at midnight

Projectile Motion

U.S.S. Missouri firing 16-inch shells - J.D. Rogers

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Starting today with math quiz

Credit equivalent to four i>clicker questions(points independent of correctness)

Please write your name on quiz booklet and scantron. Include student ID on scantron. Turn both in.

To earn quiz credit, answer the following question – who won Saturday’s game?A) U-MB) Miami of Ohio

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Questions concerning today’s youtube videos?

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Projectile motion• Two balls begin at rest at the same height (but

not at the same place). At time t = 0, one of them (A) is simply released, and the other one (B) is launched straight horizontally. (After t = 0, both of them are then freely falling.) Which one hits first? Ignore air resistance.

A. Ball A hits the ground first

B. Ball B hits the ground first

C. They both hit the ground simultaneously

D. The answer depends on which one is heavier

Let’s find out!

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Book problem – part 1 (based on E3.9)

A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350 s. Ignore air resistance. What is the height H of the tabletop above the floor?

A) 0.300 mB) 0.600 mC) 0.900 m D) 1.20 mE) 1.50 m H

y =y0 + v0yt+12

ayt2

0 =H + 0 −12

gt2

→ H =12

gt2 =(0.5)(9.80 m/s2 )(0.350 s)2 =0.600 mx motion irrelevant!

y

x

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Book problem – part 2 (based on E3.9)

A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350 s. Ignore air resistance. What is the horizontal distance from the edge of the table to the point where the book strikes the floor?

A) 0.385 mB) 0.458 mC) 0.658 m D) 0.758 mE) 0.858 m

H

x =x0 + v0xt+12

axt2

D =0 + v0xt+ 0 =(1.10 m/s)(0.350 s) = 0.385 m y motion irrelevant!

y

x

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Book problem - part 3 (based on E3.9)

A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350 s. Ignore air resistance. What is the speed of the book just before it reaches the floor?

A) 1.10 m/sB) 2.30 m/sC) 3.60 m/s D) 5.10 m/sE) 6.50 m/s H

vx =v0x + axt=v0x

vy =v0y + ayt=−gt

v= vx2 + vy

2 = (1.10 m/s)2 +[(9.80 m/s2 )(.350 s)]2 =3.60 m/s

y

x

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Forest Fire problem (based on P3.53)

In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. A pilot is practicing by dropping a canister or red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path 90.0 m above the ground and with a speed of 64.0 m/s (143 mi/h), at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

A) 166 mB) 202 mC) 239 mD) 274 mE) 311 m

Guardian 2008

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Drop point v

H = 90. m

DTarget

rv0 =(64.0 m/s) ̂i

Distance D isA) 166 mB) 202 mC) 239 mD) 274 mE) 311 m

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Drop point v

H = 90. m

DTarget

rv0 =(64.0 m/s) ̂i

Similar to book problem:

Need drop time t such that H =12

gt2

→ t =2H

g=

2(90.0 m)

9.8 m/s2= 4.28 s

→ D = v0xt = (64.0 m/s)(4.3 s) = 274. m

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Tilted Rocket problem – part 1 (based on P3.47)

A test rocket is launched by accelerating it along a 200.0-m incline at 1.25 m/s2 starting from rest at point A on the diagram. The incline rises at 35.0° above the horizontal, and at the instant the rocket leaves the incline, its engines turn off and it is subject only to gravity (ignore air resistance).What is the maximum height above the ground that the rocket Reaches?

A) 67 mB) 94 mC) 123 mD) 156 mE) 199 m

x0 = (200.0 m) cos(35.0°) = 163.8 m

y0 = (200.0 m) sin(35.0°) = 114.7 m

(x0, y0)

v02 =2as

→ v0 = 2(1.25 m/s2 )(200. m) =22.4 m/sv0

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(x0, y0)

v0

R

H

vy2 =0 =v0y

2 + 2ay(y−y0 ) =v0y

2 −2g(H −y0 )At the top:

→ H − y0 =v0y

2

2g→ H = y0 +

v0y

2

2g

→ H = 114.7 m +

[(22.4 m/s)(sin(35o)]2

(2)(9.8 m/s2 )= 123. m