Isibonelo sokunyakaza okuqonde phezulu

3 Izibonelo zemibuzo yokunyakaza eqonde phezulu

1. Ibhola liphonswa phezulu phezulu ngesivinini sokuqala esingu-20 m/s. Thola ukuphakama okuphezulu kwebhola. g = 10 m/s²
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Ekuhambeni okuqonde phezulu, lapho into inyukela phezulu, iyancipha, futhi uma ihlehla emuva, ishesha. Ngakho-ke, ukunyakaza okuqonde phezulu nakho kuyisibonelo se-GLBB.
Ifomula ye-GLBB :
v_t =v_o + ku-
s = v_o t + ½ ku²
v_t² =v_o² + ama-ekseli ama-2
Ifomula ye-GLBB engenhla ishintshiwe futhi ilungiswe ukuze ivumelane nezimo zokunyakaza okuqonde phezulu, ngamanothi amaningana.

Ifomula Yokunyakaza Okuqondile Okuya Phezulu :
v_t =v_o + gt
h = v_o t + ½ gt²
v_t² =v_o² + 2 gh
Incazelo: v_t = isivinini sokugcina, v_o = ijubane lokuqala, g = ukusheshisa ngenxa yamandla adonsela phansi, t = isikhawu sesikhathi, h = ukuphakama.
Catatan:
I-Pertama, ekuxazululeni izinkinga zokunyakaza okuqonde phezulu, inani levektha eliqondiswe phezulu linikezwa uphawu oluhle, inani levektha eliqondiswe phansi linikezwa uphawu olubi.
I-Kedua, uma indawo yokugcina yento ingaphezu kwendawo yokuqala (indawo yokuqala iyindawo yokubhekisela) khona-ke ukufuduka (h) kwento kukuhle. Ngokuphambene nalokho, uma indawo yokugcina ingaphansi kwendawo yokuqala khona-ke ukufuduka kwento kukubi.
I-Ketiga, ekuphakameni okuphezulu, into imile okwesikhashana ngaphambi kokushintsha isiqondiso, ngakho-ke ijubane lento = 0.

Kuyaziwa ukuthi:
v_o = 20 m/s (kuhle ngoba isiqondiso sejubane lokuqala siphezulu = into iphonswe phezulu), g = – 10 m/s² (okubi ngoba isiqondiso sokusheshisa kwamandla adonsela phansi sihlala sibheke phansi, v_t = 0 (ekuphakameni okuphezulu, into iphumule okwesikhashana)
Kubuziwe:
Ukuphakama okuphezulu (h) ?
Impendulo:

2. Imabula iphonswa phezulu isuka esakhiweni esingamamitha angu-100 ngaphezu komhlabathi ngesivinini sokuqala esingu-20 m/s. Nquma (a) isikhathi esidingekayo ukuze ufike phansi (b) isivinini semabula uma ifika phansi. g = 10 m/s²
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Kuyaziwa ukuthi:
h = -100 amamitha
v_o = 20m/s
g = -10 m/s²
Kubuziwe:
(a) isikhawu sesikhathi (t)
(b) isivinini sokugcina (v)_t)
Impendulo:
(a) isikhawu sesikhathi (t)
Njengoba kunikezwe i-h = -100 amamitha (i-negative ngoba indawo yokugcina yemabula ingaphansi kwendawo yokuqala yemabula), v_o = 20 m/s (kuhle ngoba isiqondiso sejubane lokuqala siphezulu noma isiqondiso sokunyakaza kokuqala siphezulu), g = -10 m/s² (okubi ngoba isiqondiso sokusheshisa kwamandla adonsela phansi siphansi).

Isikhathi asikwazi ukuba nenani elibi, ngakho-ke u-t uyasetshenziswa.₂ = imizuzwana engu-6,9.

(b) Isivinini sokugcina
Njengoba kunikezwe u-h, v_o kanye no-g, kubuzwa u-v_t, ngakho-ke sebenzisa ifomula yesithathu.

3. Ibhola A liphonswa phezulu liqonde phezulu ngesivinini esingu-10 ms ^-1 . Ngemva kwesekhondi elilodwa, kusukela endaweni efanayo, ibhola B liphonswa phezulu liqonde phezulu endleleni efanayo ngesivinini esingu-25 ms ^-1 . Ukuphakama okufinyelelwa yibhola B lapho lihlangana nebhola A...

A. 0,20 m

B. 4,80 m

C. 5,00 m

D. 5,20 m

E. 31,25 m

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Ekuxazululeni izinkinga zokunyakaza okuqonde phezulu , inani levektha eliqondiswe phezulu linikezwa uphawu oluhle, inani levektha eliqondiswe phansi linikezwa uphawu olubi.

Kuyaziwa ukuthi:

Ijubane lokuqala (v _o ) lebhola A = 10 m/s

Isikhathi esiphakathi (t) lapho ibhola u-A lisemoyeni = x

Ijubane lokuqala (v _o ) lebhola B = 25 m/s

Isikhathi esiphakathi (t) lapho ibhola B lisemoyeni = x – 1

Ukusheshisa ngenxa yamandla adonsela phansi (g) = -10 m/s ² (isiqondiso sokusheshisa ngenxa yamandla adonsela phansi siphansi ngakho-ke sinophawu olubi)

Umbuzo: Ukuphakama okufinyelelwe yibhola B uma lihlangana nebhola A (h)

Ubuningi obukhona yi-initial velocity (v _o ), i-gravitational acceleration (g), i-height (h) kanye ne-time interval (t) ngakho-ke ifomula esetshenzisiwe yile:

h = vot ₊ ½ gt ²

Ukuze kuhlangane, ukuphakama kwamabhola womabili kumele kufane.

h _A = h _B

i _- vot + ½ gt ² = i _- vot + ½ gt ²

10x + ½ (-10) x ² = 25 (x-1) + ½ (-10) (x-1) ²

10x – 5x ² = 25 (x-1) – 5 (x-1) ²

10x – 5x ² = 25x – 25 – 5 (x ² -2x+1)

10x – 5x ² = 25x – 25 – 5x ² + 10x – 5

10x – 5x ² – 25x + 25 + 5x ² – 10x + 5 = 0

– 5x ² + 5x ² + 10x – 25x – 10x + 25 + 5 = 0

10x – 25x – 10x + 25 + 5 = 0

– 25x + 25 + 5 = 0

– 25x + 30 = 0

– 25x = – 30

x = -30/-25

x = 1,2 imizuzwana

Isikhathi lapho ibhola u-A likhona emoyeni ngaphambi kokuhlangana nebhola u-B = imizuzwana eyi-1,2.

Isikhathi lapho ibhola B lisemoyeni ngaphambi kokuhlangana nebhola A = imizuzwana eyi-1,2 – umzuzwana o-1 = imizuzwana eyi-0,2.

Ukuphakama okufinyelelwe yibhola A uma lihlangana nebhola B (h):

h = vot ₊ ½ gt ² = (10)(1,2) + 1/2 (-10)(1,2) ² = 12 – 5(1,44) = 12 – 7,2 = 4,8 amamitha

Ukuphakama okufinyelelwe yibhola B uma lihlangana nebhola A (h):

h = vot ₊ ½ gt ² = (25)(0,2) + 1/2 (-10)(0,2) ² = 5 – 5(0,04) = 5 – 0,2 = 4,8 amamitha

Impendulo efanele ingu-B.

Imibuzo mayelana nokunyakaza okuqonde phezulu

1. Ibhola liphonswa phezulu phezulu ngesivinini sokuqala esingu-10 m/s. Thola ukuphakama okuphezulu kwebhola. g = 10 m/s²
Impendulo:
h = amamitha angu-5
2. Imabula iphonswa phezulu isuka esakhiweni esingamamitha angu-50 ngaphezu komhlabathi ngesivinini sokuqala esingu-5 m/s. Nquma (a) isikhathi esidingekayo ukuze ufike phansi (b) isivinini semabula uma ifika phansi. g = 10 m/s²
Impendulo:
(a) t = imizuzwana engu-3,7 (b) v_t = 32m/s

[IsiNgisi: Ukunyakaza okuphezulu nokuphansi ekuwa okukhululekile – izinkinga nezixazululo ]