[CE Board Exam Review] Dynamics - Rectilinear Motion (Free Falling Bodies)

This lecture is a review style discussion with brief introduction to concepts, important formulas, and mainly focuses in the application to civil engineering board exam problems.
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[@Jdoczterz]

I'm actually a reviewee for the CE boards and the person who explained dynamics to us in the RC eh meh, iykwim... Kaya, I'm really thankful I landed on your video. You've helped me a lot po, Sir! Auto-sub talaga ako sa mga ganitong lodi mag-explain, thank you for the crystal clear explanation, Engineer!

[@johnandreimaranan2462]

Nakapasa poba?

[@rencesalgado1106]

Almost perfect score yung quiz ko sa physics, thank you so muchhh sir, napakalaking tulong ng mga videos mo!

[@scarletwitch6059]

37:18 bat isang value lng po lumalabas sa calcu ko pag shift solve huhu, yung negative value lng po lumabas wala pong 9.829 sec please help meee

[@jordanc.837]

dba 0 ang initial velocity pag bumaba from 186.055 meters to the ground tapos positive 9.81 ang acceleration kasi going downward due to gravity kaya positive pag downward then negative pag upward, therefor formula is S= 0 + 1/2 at^2 ,which is 186.055= 0 + 1/2(9.81)(t)^2 simplify the formula, we have t= squaroot of 186.055/4.905 = 6.159 + yung 3.670 from the initial point of the ball to maximum height kaya 6.519 + 3.670 = 9.829s no need to shiftsolve just analyze carefully and understand the situation thank me later

[@kenjohnmac4841]

equation po gamitin mo pag 2nd degree ang equation. pag shift solve Isa lang pinapakita Nyan.

[@beefsteaktv6177]

Yung problem 16 po which is it take to reach the ground. Iba po yung nakukuha na answer pag sinolve niyu po yung time to take from point 120 to highest peak then time from 186.055 to point 0 whcih makukuha mo pa din ang time pag i-add yung dalawa ang sagot po is 9.78 seconds. (Datum ko po is ground) Time =0 S1=120 V1=36 T2=3.67 V2=0 S2=186.055 T3=3.67+6.11(TRAVEL TIME FROM S2 to S3)=9.78 V3=60.419m/s S3=0(DATUM)

[@jordanc.837]

dba 0 ang initial velocity pag bumaba from 186.055 meters to the ground tapos positive 9.81 ang acceleration kasi going downward due to gravity kaya positive pag downward then negative pag upward, therefor formula is S= 0 + 1/2 at^2 ,which is 186.055= 0 + 1/2(9.81)(t)^2 simplify the formula, we have t= squaroot of 186.055/4.905 = 6.159 + yung 3.670 from the initial point of the ball to maximum height kaya 6.519 + 3.670 = 9.829s no need to shiftsolve just analyze carefully and understand the situation thank me later

[@rdcsantosksa]

thank you engineer

[@naulty1827]

Pwede nyo iderive ung formula sa 37:26 para sa t3 = (time total) = V1/g + √2s/g bali 36/9.81 + √2(186.055)/9.81

[@emmanlazo9529]

Pwede pong pa explain how? Thank you

[@echo6935]

Sir, ask ko po kung sa 37:41 ay dalawang positive na time ang lumabas, ano po pipiliin?salamat po

[@missRara-tk9dd]

Sir sa 37:26 pano po makuha yung isang t3 sa calcu?😭

[@missRara-tk9dd]

Yung t3= 9.829 how po yun?😭

[@jenicolayog8924]

@@missRara-tk9dd sir gamit ka po ng shift solve sa calcu, let t3 be the x sa pag input po

[@jaylesterdapat7421]

since quadratic equation po. yung sagot ay positive and negative .. if mag shift solve kana kong gusto mo makoha ang positive value set ka ng positive value if gusto mong makoha ang negative value set ka nga negative value like -10

[@rivelynmartinez7602]

Thank you sir.

[@wanhandredwan]

okay po

[@carlarnoldp.bernabe2102]

gaganahan ako mag aral kung ganto teacher

[@jannicadelosreyes6177]

Di ko talaga alam pano nakuha yung postive na t3 sa 37:50 😢

[@magelynmartin2945]

Caltech po. Mode(5), #3, input niyo po 1/2(-9.81) para sa (a) 36 para sa (b) then 120 para sa (c). Press equal.

[@trunxsvincentdoligol502]

@@magelynmartin2945 bakit po positive 120 ang ginamit na sa given po nang equation ay -120 po nan doon. salamat po

[@cruelsummer3021]

@@trunxsvincentdoligol502hi! a+b=c kasi ng format sa calcu

[@ejezzjes3928]

Sir pano po naging 9.825 yung sa t3

[@reignewellcatoy6579]

roots po nung quadratic equation