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NAME:

SOLUTION

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PHYS207-19S

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Quiz 2

1 March 2019

g  9.8 m / s2  32 ft/ s2  40_1. Apackageisdroppedfromahelicoptermovingupwardat15m/s.Ifittakes5secondsbefore the package strikes the ground, how high above the ground was the package when it was releasedandwhatisitsmaximumheightifairresistanceisnegligible?(10points) 

Weknowthat y f  0 , t f  16s and v0  15 m / s .Thepointwhenthepackageleavesthe helicopter,wesett=0anditisinfreefallsuchthat a y   g .

v   gt  v 0 

y 

1 2 gt  v 0t  y 0  2

Maximumheightoccurswhen v  0 ,so, v0  gt .

1 2 gt  v0 t f  y0 2 f 1 2 2 y0  9.8 m / s  5s   15 m / s 5s   2 y0  43.75 m

0  y0 

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1 2 gt  v0 t f 2 f

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Themaximumheightoccurswhen v = 0 . 2

v0  gtmax

ymax

1 v  v    g  0   v0  0   y0 2 g  g

v2 15 m / s   0  y0   43.75 m 2g 2  9.8 m / s2  2

ymax

y max  55.24 m      

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NAME:

SOLUTION

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41_1. Apackageisdroppedfromahelicoptermovingupwardat12m/s.Ifittakes7secondsbefore the package strikes the ground, how high above the ground was the package when it was releasedandwhatisitsmaximumheightifairresistanceisnegligible?(10points) 

Weknowthat y f  0 , t f  16s and v0  15 m / s .Thepointwhenthepackageleavesthe helicopter,wesett=0anditisinfreefallsuchthat a y   g .

v   gt  v 0 

1 2 gt  v 0t  y 0  2

Maximumheightoccurswhen v  0 ,so, v0  gt .

y0 

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y 

1 2 gt  v0 t f 2 f

1 0   gt 2f  v0t f  y 0 2 1 2 y0   9.8 m / s 2   7 s   12 m / s  7 s   2 y0  156 m

Themaximumheightoccurswhen v = 0 . 2

v0  gtmax

ymax

v  1 v    g  0   v0  0   y0 2 g  g

12 m / s  v20  y0   156 m 2 2g 2  9.8 m / s  2

ymax 

ymax  163m  

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NAME:

SOLUTION

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42_1. Withwhatminimuminitialvelocitydoyouneedtothrowaballupwardtoreachaheightof 9.0m?(6points) 

Maximumheightoccurswhen v  0 .Settingthelaunchheightat y  0 ,weneedtofind v0 .

a  g

 v   gt  v 0  1 y   gt 2 + v0t  y 0 2

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With v  0 , tmax 

v0 . g 2

ymax

1 v  v    g  0  + v0  0  2 g  g 



ymax 

v0  2 gymax  2  9.8 m / s  9.0 m 

v02 2g 

v 0  13.2 m / s  42_2

Specifythevelocitiesandaccelerationsofthepointsnotedonthegraphaspositive,negative orzero(4points)

 Time A B C D   

Velocity 0 ‐ 0 +

Acceleration ‐ 0 + +

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NAME:

SOLUTION

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43_1. TheobservationdeckoftheEmpireStateBuildingis112mabovetheground.Atwhatinitial velocity(specifyupordown)doyouneedtothrowthepennytohaveitstrikethegroundin halfthetimeasjustlettingitdrop?(6points) 

Thereareseveralwaystosolvethis,oneistofindthetimeittakestofallwithnoinitialvelocity, halvethatandsubstitutebackintotheeqnofmotiontofind v0 .

1 y   gt 2  y 0 2 2

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 t ff 1  t ff  0   g    v0  2  2  2 v0 

t ff 

   y0 

2  ygr  y0  g 

v0 

 4.8s

2y 1 gt ff  0  4 t ff

2 112 m  1 9.8m / s2   4.8s    4 4.8s v0  35m / s

 43_2

Specifythevelocitiesandaccelerationsofthepointsnotedonthegraphaspositive,negative orzero(4points)

 Time A B C D 

Velocity 0 + + 0

Acceleration + + 0 +

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