I-physics

1. Ibhola elikhahlelwayo lishiya umhlabathi nge-engeli engu-θ = 30 ^o liye endaweni evundlile ngesivinini sokuqala esingu-14 m/s. Bala ijubane lokugcina ngaphambi kokuba ibhola lishaye phansi.

Kwaziwa:

I-engeli ( θ ) = 3 0 ^o

Ijubane lokuqala (v _o ) = 14 m/s

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Okufunwayo: Ijubane lokugcina ngaphambi kokuba ibhola lishaye phansi

Isixazululo:

Ingxenye evundlile yejubane lokuqala:

v _inkabi = v _o cos θ = (14 m/s)(cos 30 ^o ) = (14 m/s)(0.5 √ 3 ) = 7 √ 3 m/s

Ingxenye eqondile yejubane lokuqala:

v _oy = v _o isono θ = (14 m/s)(isono 30 ^o ) = (14 m/s)(0.5) = 7 m/s

Ijubane lokugcina esiqondisweni esiqondile

Khetha indlela eya phezulu njengeyakhayo kanye nendlela eya phansi njengeyakhayo.

Kwaziwa:

Ijubane lokuqala (v _o ) = 7 m/s (elihle phezulu)

Ukusheshisa kwamandla adonsela phansi (g) = – 10 m/s ² (ukwehla okungekuhle)

Ukuphakama (h) = 0 (into ibuyele endaweni yokuqala)

Okufunwayo: Ijubane lokugcina (v _t )

Isixazululo:

v _t ² = v _o ² + 2 gh = 7 ² + 2(-10) (0) = 49 – 0 = 49

v _t = √49 = 7 m/s

Ijubane lokugcina endaweni evundlile

Ijubane lokuqala ohlangothini oluvundlile lingu- 7 √ 3 m/s. Ijubane aliguquguquki ukuze ijubane lokugcina lifane nejubane lokuqala.

Ijubane lokugcina ngaphambi kokuba into ishaye phansi

2. Umzimba uphakanyiswa phezulu nge-engeli engu-30 ^o kanti ukuphakama kwawo kungama-5 metres ukusuka esakhiweni. Ijubane lawo lokuqala lingu-10 m/s. Bala ijubane lokugcina ngaphambi kokuba into ishaye phansi! Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ².

Kwaziwa:

I-engeli ( θ ) = 30 ^o

Ukuphakama kokuqala (h _o ) = amamitha ayi-5

Ijubane lokuqala (v _o ) = 10 m/s

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Okufunwayo: Ijubane lokugcina

Isixazululo:

Ingxenye evundlile yejubane lokuqala:

v _inkabi = v _o cos θ = (10 m/s)(cos 30 ^o ) = (10 m/s)(0.5 √ 3 ) = 5 √ 3 m/s

Ingxenye eqondile yejubane lokuqala:

v _oy = v _o isono θ = (10 m/s)(isono 30 ^o ) = (10 m/s)(0.5) = 5 m/s

Ijubane lokugcina esiqondisweni esiqondile

Kwaziwa:

Ijubane lokuqala (v _o ) = 5 m/s (elihle phezulu)

Ukusheshisa kwamandla adonsela phansi (g) = – 10 m/s ² (ukwehla okungekuhle)

Ukuphakama (h) = -5 m ( okungekuhle ngoba umhlabathi ungaphansi kokuphakama kokuqala)

Okufunwayo: Ijubane lokugcina (v _t )

Isixazululo:

v _t ² = v _o ² + 2 gh = 5 ² + 2(-10)(-5) = 25 + 100 = 125

v _t = √125 m/s

Ijubane lokugcina endaweni evundlile

Ijubane lokugcina endaweni evundlile lingu- 5 √3 m/s.

Ijubane lokugcina

3. Ibhola elincane lilinganiswa ngokuvundlile ngesivinini sokuqala v _o = 8 m/s ukusuka esakhiweni esingamamitha ayi-12 ukuphakama. Bala isivinini sokugcina ngaphambi kokuba ibhola lishaye phansi ! Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ²

Kwaziwa:

Ukuphakama (h) = amamitha angu-12

Ijubane lokuqala (v _o ) = 8 m/s

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Okufunwayo: Ijubane lokugcina (v _t )

Isixazululo:

Ingxenye evundlile yejubane lokuqala:

v _ox = v _o = 8 m/s

Ingxenye eqondile yejubane lokuqala:

v _oy = 0 m/s

Ijubane lokugcina esiqondisweni esiqondile

kubalwe kusetshenziswa i-equation ka ukunyakaza kokuwa kwamahhala.

Kwaziwa:

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Ukuphakama (h) = 12 m

Okufunwayo: Ijubane lokugcina (v _t )

Isixazululo:

v _t ² = 2 gh = 2(10)(12) = 240

v _t = √240 m/s

Ijubane lokugcina endaweni evundlile

Ijubane lokuqala ohlangothini oluvundlile lingu-8 m/s. Ijubane aliguquguquki ukuze ijubane lokuqala lilingane nejubane lokugcina. Ngakho ijubane lokugcina ohlangothini oluvundlile lingu-8 m/s.

Ijubane lokugcina

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1. Xazulula ijubane lokuqala libe izingxenye ezivundlile neziqondile
2. Nquma ukufuduka okuvundlile
3. Nquma ukuphakama okuphezulu
4. Nquma isikhathi esinqunyiwe
5. Nquma indawo yento
6. Nquma ijubane lokugcina

Funda kabanzi

Izinkinga ezixazululiwe ekuhambeni kwe-projectile - thola indawo yento

1. Umzimba uphakanyiswa phezulu nge-engeli engu-60 ^o uye endaweni evundlile ngesivinini sokuqala esingu-12 m/s. Thola indawo yento ngemva kokuhamba umzuzwana owodwa! Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ².

Kwaziwa:

I-engeli ( θ ) = 60 ^o

Ijubane lokuqala (v _o ) = 12 m/s

Isikhawu sesikhathi (t) = umzuzwana o-1

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Okufunwayo: Indawo yento ngemva kokuhamba umzuzwana owodwa

Isixazululo:

Ingxenye evundlile yejubane lokuqala:

v _inkabi = v _o cos θ = (12 m/s)(cos 60 ^o ) = (12 m/s)(0.5) = 6 m/s

Ingxenye eqondile yejubane lokuqala:

v _oy = v _o isono θ = (12 m/s)(isono 60 ^o ) = (12 m/s)(0.5 √ 3 ) = 6 √ 3 m/s

Indawo yento endaweni evundlile:

Kwaziwa:

Ingxenye evundlile yejubane (v _x ) = 6 m/s

Isikhawu sesikhathi (t) = umzuzwana o-1

Okufunwayo: ububanzi obuvundlile (x)

Isixazululo:

Amamitha ayi-6/umzuzwana kusho ukuthi ibhola lihamba lize lifike kumamitha ayi-6 njalo ngomzuzwana owodwa. Ibanga lebhola ngemva kokuhamba umzuzwana owodwa lingamamitha ayi-6. Ngakho-ke indawo yebhola ohlangothini oluvundlile ingamamitha ayi-6.

Indawo yento esiqondisweni esiqondile:

Khetha indlela eya phezulu njengeyakhayo kanye nendlela eya phansi njengeyakhayo.

Kwaziwa:

Ijubane lokuqala (v _o ) = 6 √ 3 m/s (elihle phezulu)

Isikhawu sesikhathi (t) = umzuzwana o-1

Ukusheshisa kwamandla adonsela phansi (g) = -10 m/s ² (ukwehla okungekuhle)

Kufunwa: ukuphakama ngemva kokuhamba umzuzwana owodwa

Isixazululo:

h = v_o t + 1/2 gt² = ((6√3(1) + 1/2 (-10)(1²) = 6√3 + (-5)(1) = 6√3 – 5 = 6(1.7) – 5 = 10.2 – 5 = 5.2 amamitha.

Indawo yento ngemva kokuhamba isekhondi eli-1:

Ukuhamba okuvundlile (x) = amamitha ayi-6

Ukuhamba okuqondile (y) = amamitha angu-5.2

2. Umzimba uphakanyiswa phezulu nge-engeli engu-30 ^o uye endaweni evundlile ukusuka esakhiweni esingamamitha angu-20 ukuphakama. Isivinini sawo sokuqala singama-50 m/s. Bala ukufuduka okuqondile ngemva kokuba umzimba unyakaze umzuzwana owodwa! Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ².

Kwaziwa:

I-engeli ( θ ) = 30 ^o

Ukuphakama kokuqala (h _o ) = amamitha ayi-20

Ijubane lokuqala (v _o ) = 50 m/s

Isikhawu sesikhathi (t) = umzuzwana o-1

Ukusheshisa kwamandla adonsela phansi (g) = 1 0 m/s ²

Okufunwayo: Ukuphakama (h)

Isixazululo:

Ingxenye eqondile yejubane lokuqala:

v _oy = v _o isono θ = (50 m/s)(isono 30 ^o ) = (50 m/s)(0.5) = 25 m/s

Ukuphakama:

Khetha indlela eya phezulu njengeyakhayo kanye nendlela eya phansi njengeyakhayo.

Kwaziwa:

Ijubane lokuqala (v _o ) = 25 m/s (elihle phezulu)

Isikhawu sesikhathi (t) = umzuzwana o-1

Ukusheshisa kwamandla adonsela phansi (g) = -1 0 m/s ² (ukwehla okungekuhle)

Okufunwayo: Ukuphakama (h)

Isixazululo:

h = vot ₊ 1/2 gt ^{​​​​2} = (25)(1) + 1/2 (-10)(1 ² ) = 25 + (-5)(1) = 25 – 5 = amamitha angu-20.

Ukuphakama komzimba ngemva kokuhamba umzuzwana owodwa kungamamitha angama-20 ngaphezu kwalapho umzimba ubekwe khona noma amamitha angama-40 ngaphezu komhlabathi.

3. Ibhola elincane liphakanyiswa ngokuvundlile ngesivinini sokuqala v _o = 10 m/s ukusuka esakhiweni esingamamitha ayi-10 ukuphakama. Bala ukufuduka kwebhola ngemva kokuhamba umzuzwana owodwa ! Ukusheshisa kwamandla adonsela phansi yi-10 m/s ²

Kwaziwa:

Ukuphakama kokuqala (h) = amamitha ayi-10

Ijubane lokuqala (v _o ) = 10 m/s

Isikhawu sesikhathi (t) = umzuzwana o-1

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Okufunwayo: Indawo yebhola ngemva kokuhamba umzuzwana owodwa!

Isixazululo:

Ukufuduka okuvundlile:

Kwaziwa:

Ingxenye evundlile yejubane (v _x ) = 10 m/s

Isikhawu sesikhathi (t) = umzuzwana o-1

Okufunwayo: Indawo yento

Isixazululo:

Amamitha ayi-10/umzuzwana kusho ukuthi into ihamba ibanga elingafika kumamitha ayi-10 njalo ngomzuzwana owodwa. Ukufuduka ngemva kokuhamba umzuzwana owodwa kungamamitha ayi-10. Ngakho ukufuduka okuvundlile kungamamitha ayi-10.

Ukufuduka okuqondile:

Kubalwa njengokunyakaza kokuwa okukhululekile.

Kwaziwa:

Isikhawu sesikhathi (t) = umzuzwana o-1

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Okufunwayo: Ukuphakama ngemva kokuhamba umzuzwana owodwa (h)

Isixazululo:

h = 1/2 gt ^{​​​​2} = 1/2 (10)(1 ² ) = (5)(1) = amamitha ama-5.

Ngemva komzuzwana owodwa, into iwela amamitha angu-5. Ukuphakama ngaphezu komhlabathi = amamitha ayi-10 - amamitha angu-5 = amamitha angu-5.

Indawo yento ngemva kokuhamba umzuzwana owodwa:

Indawo yento endaweni evundlile (x) = amamitha ayi-10

Indawo yento eqonde ngqo (y) = amamitha angu-5

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1. Xazulula ijubane lokuqala libe izingxenye ezivundlile neziqondile
2. Nquma ukufuduka okuvundlile
3. Nquma ukuphakama okuphezulu
4. Nquma isikhathi esinqunyiwe
5. Nquma indawo yento
6. Nquma ijubane lokugcina

Funda kabanzi

Izinkinga ezixazululiwe ekunyakazeni kwe-projectile - nquma isikhathi sokuphumula

1. Ibhola elikhahlelwayo lishiya umhlabathi nge-engeli engu-θ = 30 ^o liye endaweni evundlile ngesivinini sokuqala esingu-10 m/s. Bala isikhathi sokuphumula ukuze ufinyelele ukuphakama okuphezulu! Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ².

Kwaziwa:

I-engeli (θ) = 3 0 ^o

Ijubane lokuqala (v _o ) = 10 m/s

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Okufunwayo: Isikhathi sokufinyelela ukuphakama okuphezulu

Isixazululo:

Ingxenye eqondile yejubane lokuqala:

v _oy = v _o isono θ = (10 m/s)(isono 30 ^o ) = (10 m/s)(0.5) = 5 m/s

Isikhathi sokufinyelela ukuphakama okuphezulu sinqunywa yi- equation yokunyakaza okuqondile . Khetha isiqondiso saphezulu njengesihle kanye nesiqondiso saphansi njengesibi.

Kwaziwa:

Ijubane lokuqala (v _o ) = 5 m/s (elihle phezulu)

Ukusheshisa kwamandla adonsela phansi (g) = – 10 m/s ² (ukwehla okungekuhle)

Ijubane lokugcina ekuphakameni okuphezulu (v _t ) = 0

Okufunwayo: isikhathi sokuphumula (t)

Isixazululo:

v _t = vo _o + g t

0 = 5 + (-10)t

0 = 5 – 10 t

5 = 10 t

t = 5/10 = 0.5 imizuzwana

2. Umzimba uphakanyiswa phezulu nge-engeli engu-30 ^o uye endaweni evundlile ngesivinini sokuqala esingu-30 m/s. Bala isikhathi sokundiza! Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ².

Kwaziwa:

I-engeli (θ) = 3 0 ^o

Ijubane lokuqala (v _o ) = 8 m/s

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Kufunwa: Isikhathi esinqunyiwe ngaphambi kokuba umzimba ushaye phansi

Isixazululo:

Ingxenye eqondile yejubane lokuqala:

v _oy = v _o isono θ = (8 m/s)(isono 30 ^o ) = (8 m/s)(0.5) = 4 m/s

Siqala ngokubala isikhathi esiphakathi ukuze sifinyelele ukuphakama okuphezulu sisebenzisa i-equation yokunyakaza okuqondile.

Khetha indlela eya phezulu njengeyakhayo kanye nendlela eya phansi njengeyakhayo.

Kwaziwa:

Ijubane lokuqala (v _o ) = 4 m/s (elihle phezulu)

Ukusheshisa kwamandla adonsela phansi (g) = – 10 m/s ² (ukwehla okungekuhle)

Ijubane lokugcina ekuphakameni okuphezulu (v _t ) = 0

Okufunwayo: Isikhathi esiphakathi (t)

Isixazululo:

v _t = vo _o + g t

0 = 4 + (-10)t

0 = 4 – 10 t

4 = 10 t

t = 4/10 = 0,4 imizuzwana

Isikhathi sokufinyelela ukuphakama okuphezulu singamasekhondi angu-0.4.

Isikhathi emoyeni singamasekhondi angu-2 x 0.4 = 0.8.

3. Umzimba uphakanyiswa ubheke phezulu nge-engeli engu-30 ^o kanti ukuphakama kwawo kungama-10 metres ukusuka esakhiweni. Ijubane lawo lokuqala lingu-40 m/s. Kuthatha isikhathi esingakanani ukuthi umzimba ufike phansi? Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ².

Kwaziwa:

I-engeli (θ) = 3 0 ^o

Ukuphakama kokuqala (h _o ) = amamitha ayi-10

Ijubane lokuqala (v _o ) = 40 m/s

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Kufunwa: Isikhathi emoyeni (t)

Isixazululo:

Ingxenye eqondile yejubane lokuqala:

v _oy = v _o isono θ = (40 m/s)(isono 30 ^o ) = (40 m/s)(0.5) = 20 m/s

Siqala ngokubala isikhathi esiphakathi ukuze sifinyelele ukuphakama okuphezulu sisebenzisa i-equation yokunyakaza okuqondile.

Khetha indlela eya phezulu njengeyakhayo kanye nendlela eya phansi njengeyakhayo.

Kwaziwa:

Ijubane lokuqala (v _o ) = 20 m/s (elihle phezulu)

Ukusheshisa kwamandla adonsela phansi (g) = – 10 m/s ² (ukwehla okungekuhle)

Ijubane lokugcina esicongweni (v _t ) = 0

Okufunwayo: Isikhathi esiphakathi (t)

Isixazululo:

v _t = vo _o + g t

0 = 20 + (-10)t

0 = 20 – 10 t

20 = 10 t

t = 20/10 = imizuzwana emi-2

Isikhathi emoyeni = 2 x 2 imizuzwana = 4 imizuzwana.

Into ingamamitha ayi-10 ngaphezu komhlabathi. Imizuzwana emi-4 yisikhathi sokufinyelela endaweni ehambisana nendawo yokuqala. Ibhola lisahamba liye phansi.

Isikhathi sokufinyelela phansi sibalwa kusetshenziswa i-equation yokunyakaza kokuwa okukhululekile.

Kwaziwa:

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Ukuphakama (h) = amamitha ayi-10

Okufunwayo: Isikhathi esiphakathi (t)

Isixazululo:

h = 1/2 gt ^{​​​​2}

10 = 1/2 (10) t ²

10 = 5 t ²

t ² = 10/5 = 2

t = √2 = 1.4 imizuzwana

Isikhawu sesikhathi = imizuzwana eyi-1.4.

Isikhawu sesikhathi esiphelele = imizuzwana emi-4 + imizuzwana eyi-1.4 = imizuzwana emi-5.4.

4. Ibhola elincane liphakanyiswa ngokuvundlile ngesivinini sokuqala v _o = 15 m/s ukusuka esakhiweni esingamamitha ama-5 ukuphakama. Bala isikhathi emoyeni ! Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ²

Kwaziwa:

Ukuphakama (h) = amamitha ayi-5

Ijubane lokuqala (v _o ) = 15 m/s

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Kufunwa: Isikhathi emoyeni (t)

Isixazululo:

Isikhathi emoyeni sibalwa kusetshenziswa isibalo sokunyakaza okuwa ngokukhululeka.

Kwaziwa:

Ukuphakama (h) = amamitha ayi-5

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Okufunwayo: Isikhathi esiphakathi (t)

Isixazululo:

h = 1/2 gt ^{​​​​2}

5 = 1/2 (10) t ²

5 = 5 t ²

t ² = 5/5 = 1

t = √1 = 1 umzuzwana

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[wpdm_package id='536′]

1. Xazulula ijubane lokuqala libe izingxenye ezivundlile neziqondile
2. Nquma ukufuduka okuvundlile
3. Nquma ukuphakama okuphezulu
4. Nquma isikhathi esinqunyiwe
5. Nquma indawo yezinto
6. Nquma ijubane lokugcina

Funda kabanzi

Izinkinga ezixazululiwe ekuhambeni kwe-projectile - thola ukuphakama okuphezulu

1. Ibhola elikhahlelwayo lishiya umhlabathi nge-engeli engu-θ = 60 ^o kanti elivundlile linesivinini sokuqala esingu-10 m/s. Bala ukuphakama okuphezulu! Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ².

Kwaziwa:

I-engeli (θ) = 60 ^o

Isivinini sokuqala (v _o ) = 10 m/s

Okufunwayo: Ukuphakama okuphezulu (h)

Isixazululo:

Ingxenye eqondile yejubane lokuqala:

isono 60 ^o = v _oy / v _o

v _oy = v _o isono 60 ^o = (10)(isono 60 ^o ) = (10)(0.5 √ 3 ) = 5 √ 3 m/s

Khetha indlela eya phezulu njengeyakhayo kanye nendlela eya phansi njengeyakhayo.

Kwaziwa:

Ukusheshisa kwamandla adonsela phansi (g) = -10 m/s ² (ukwehla okungekuhle)

Ingxenye eqondile yejubane lokuqala (v _oy ) = + 5 √ 3 m/s (ephezulu phezulu)

Ijubane lokugcina ekuphakameni okuphezulu (v _ty ) = 0

Okufunwayo: Ukuphakama okuphezulu (h)

Isixazululo:

v _t ² = v _o ² + 2 gh

0 ² = ( 5 √ 3) ² + 2 (-10) h

0 = 2 5 ( 3 ) – amahora angu-20

0 = 75 – 20 amahora

75 = amahora angu-20

h = 75/20

h = amamitha angu-3.75

Ukuphakama okuphezulu kungamamitha angu-3.75.

2. Umzimba uphakanyiswa ubheke phezulu nge-engeli engu-30 ^o kanti ukuphakama kwawo kungama-20 metres ukusuka esakhiweni. Isivinini sawo sokuqala singama-4 m/s. Bala ukuphakama okuphezulu! Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ².

Kwaziwa:

I-engeli (θ) = 30 ^o

Ukuphakama kokuqala (h) = amamitha angu-20

Ijubane lokuqala (v _o ) = 4 m/s

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Okufunwayo: Ukuphakama okuphezulu (h)

Isixazululo:

Ingxenye eqondile yejubane lokuqala:

isono 30 ^o = v _oy / v _o

v _oy = v _o isono 30 ^o = (4)(isono 30 ^o ) = (4)(0.5) = 2 m/s

Khetha indlela eya phezulu njengeyakhayo kanye nendlela eya phansi njengeyakhayo.

Kwaziwa:

Ukusheshisa kwamandla adonsela phansi (g) = -10 m/s ² (ukwehla okungekuhle)

Ingxenye eqondile yejubane lokuqala (v _oy ) = +2 m/s (ephezulu phezulu)

Ijubane lokugcina ekuphakameni okuphezulu (v _ty ) = 0

Okufunwayo: Ukuphakama okuphezulu

Isixazululo:

Ukuphakama okuphezulu:

v _t ² = v _o ² + 2 gh

0 ² = 2 ² + 2 (-10) amahora

0 = 4 – 20 amahora

4 = amahora angu-20

h = 4/20

h = amamitha angu-0.2

Ukuphakama okuphezulu kakhulu kungamamitha angu-0.2 + amamitha angu-20 = amamitha angu-20.2.

[wpdm_package id='528′]

[wpdm_package id='536′]

1. Xazulula ijubane lokuqala libe izingxenye ezivundlile neziqondile
2. Nquma ukufuduka okuvundlile
3. Nquma ukuphakama okuphezulu
4. Nquma isikhathi esinqunyiwe
5. Nquma indawo yezinto
6. Nquma ijubane lokugcina

Funda kabanzi

Izinkinga ezixazululiwe ekuhambeni kwe-projectile - thola ukufuduka okuvundlile

1. Ibhola elikhahlelwayo lishiya phansi nge-engeli engu-θ = 60 ^o kanti elivundlile linejubane lokuqala elingu-16 m/s. Kuzoba isikhathi esingakanani ngaphambi kokuba ibhola lishaye phansi?

Kwaziwa:

I-engeli ( θ ) = 60 ^o

Isivinini sokuqala (v _o ) = 16 m/s

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Okufunwayo: Ukufuduka okuvundlile (x)

Isixazululo:

Ingxenye evundlile yejubane lokuqala:

v _inkabi = v _o cos θ = (16 m/s)(cos 60 ^o ) = (16 m/s)(0.5) = 8 m/s

Ingxenye eqondile yejubane lokuqala:

v _oy = v _o isono θ = (16 m/s)(isono 60 ^o ) = (16 m/s)(0.5 √ 3 ) = 8 √ 3 m/s

Ukunyakaza okubonakalayo kungaqondwa ngokuhlaziya izingxenye ezivundlile neziqondile zokunyakaza ngokwehlukana. Ukunyakaza kuka-x kwenzeka ngesivinini esingaguquki kanti ukunyakaza kuka-y kwenzeka ngokusheshisa okungaguquki kwamandla adonsela phansi.

Isikhathi emoyeni

Isikhathi esihlala emoyeni sinqunywa ukunyakaza kuka-y. Siqala ngokuthola isikhathi sisebenzisa ukunyakaza kuka-y bese sisebenzisa leli nani lesikhathi kuzibalo zika-x ( isilinganiso sejubane esingaguquki ).

Khetha indlela eya phezulu njengeyakhayo kanye nendlela eya phansi njengeyakhayo.

Kwaziwa:

Ijubane lokuqala (v _o ) = 8 √ 3 m/s (v _o phezulu)

Ukusheshisa kwamandla adonsela phansi (g) = -10 m/s ² (g phansi)

Ukuphakama (h) = 0 (ibhola libuyele endaweni efanayo)

Kufunwa: Isikhathi Esisemoyeni

Isixazululo:

h = vot ₊ 1/2 gt ^{​​​​2}

0 = ( 8 √ 3) t + 1/2 (-10) t ²

0 = 8 √ 3 t – 5 t ²

8 √ 3 t = 5 t ²

8 (1.7) = 5 t

14 = 5 t

t = 14 / 5 = 2.8 imizuzwana

Ukufuduka okuvundlile

Kwaziwa:

Ijubane (v) = 8 m/s

Isikhawu sesikhathi (t) = imizuzwana eyi-2.8

Okufunwayo: Ukufuduka

Isixazululo:

x = vt = (8 m/s)(2.8 s) = 22.4 amamitha

Ukufuduka okuvundlile kungamamitha angu-22.4.

2. Umzimba uphakanyiswa ubheke phezulu nge-engeli engu-60 ^o kanti ukuphakama kwawo kuqondile kusuka esakhiweni ngamamitha angu-50. Isivinini sawo sokuqala singama-30 m/s. Bala ukufuduka okuqondile! Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ².

Kwaziwa:

I-engeli ( θ ) = 60 ^o

Ukuphakama (h) = 15 m

Isivinini sokuqala (v _o ) = 30 m/s

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Kufunwa: x

Isixazululo:

Ingxenye evundlile yejubane lokuqala ::

v _inkabi = v _o cos θ = (30 m/s)(cos 60 ^o ) = (30 m/s)(0.5) = 15 m/s

Ingxenye eqondile yejubane lokuqala:

v _oy = v _o isono θ = (30 m/s)(isono 60 ^o ) = (30 m/s)(0.5 √ 3 ) = 15 √ 3 m/s

Isikhathi emoyeni

Siqala ngokuthola isikhathi sisebenzisa ukunyakaza kuka-y bese sisebenzisa leli nani lesikhathi kuzibalo zika-x (isilinganiso sejubane eliqhubekayo). Khetha phezulu njengehle kanye nephansi njengehle.

Kwaziwa:

Ijubane lokuqala (v _o ) = 15 √ 3 m/s (elihle phezulu)

Ukusheshisa kwamandla adonsela phansi (g) = -10 m/s ² (ukwehla okungekuhle)

Ukuphakama (h) = -50 (Umhlaba ngamamitha angu-50 ngaphansi kwendawo yokuqala)

Kufunwa: t

Isixazululo:

h = vot ₊ 1/2 gt ^{​​​​2}

-50 = ( 15 √ 3 ) t + 1/2 (-10) t ²

-50 = 15 √ 3 t – 5 t ²

5 t ² – 15 √ 3 t – 50 = 0

Bala isikhathi usebenzisa le fomula:

a = 5, b = – 15 √ 3, c = – 50

Isikhathi emoyeni singamasekhondi angu-6.7.

Ukufuduka okuvundlile:

Kwaziwa:

Ijubane (v) = 15 m/s

Isikhawu sesikhathi (t) = imizuzwana eyi-6.7

Kufunwa: ukufuduka

Isixazululo:

s = vt = (15 m/s)(6.7 s) = amamitha ayi-100.5

Ukufuduka okuvundlile kungamamitha angu-100.5.

3. Ibhola elincane liphakanyiswa ngokuvundlile ngesivinini sokuqala v _o = 10 m/s ukusuka esakhiweni esingamamitha ayi-10 ukuphakama. Bala ukufuduka okuvundlile ! Ukusheshisa kwamandla adonsela phansi kungu-10 m/s ²

Kwaziwa:

Ukuphakama (h) = 10 m

Ijubane lokuqala (v _o ) = 10 m/s

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Kufunwa: x

Isixazululo:

Ingxenye evundlile yejubane lokuqala = ijubane lokuqala = 10 m/s.

Isikhathi emoyeni

Isikhathi emoyeni sibalwa kusetshenziswa i-equation yokunyakaza kokuwa kwamahhala.

Kwaziwa:

Ukusheshisa kwamandla adonsela phansi (g) = 10 m/s ²

Ukuphakama (h) = amamitha ayi-10

Kufunwa: t

Isixazululo:

h = 1/2 gt ^{​​​​2}

10 = 1/2 (10) t ²

10 = 5 t ²

t ² = 10 / 5 = 2

t = √2 = 1.4 imizuzwana

Ukufuduka okuvundlile

Ukufuduka okuvundlile kubalwa kusetshenziswa i-equation yokunyakaza ngesivinini esingaguquki.

Kwaziwa:

Ijubane (v) = 10 m/s

Isikhawu sesikhathi (t) = imizuzwana eyi-1.4

Kufunwa: x

Isixazululo:

s = vt = (10 m/s)(1.4 s) = amamitha ayi-14

Ukufuduka okuvundlile kungamamitha angu-14.

[wpdm_package id='526′]

[wpdm_package id='536′]

1. Xazulula ijubane lokuqala libe izingxenye ezivundlile neziqondile
2. Nquma ukufuduka okuvundlile
3. Nquma ukuphakama okuphezulu
4. Nquma isikhathi esinqunyiwe
5. Nquma indawo yezinto
6. Nquma ijubane lokugcina

Funda kabanzi

Izinkinga ezixazululiwe ekunyakazeni okuqoshwe phansi - ijubane lokuqala lixazululwe libe izingxenye ezivundlile neziqondile

1. Ibhola elikhahlelwe lishiya phansi nge-engeli θ = 60^o ngejubane elingu-10 m/s. Bala izingxenye zejubane lokuqala!
Kwaziwa:
I-engeli (θ) = 60^o
Ijubane lokuqala (v)_o) = 10 m/s
Kufunwa: v_ox kanye no-v_oy
Isixazululo:
Xazulula ijubane lokuqala libe yi-x component (evundlile) kanye ne-y component (evundlile).
isono θ = v_oy /v_o —–> v_oy =v_o isono θ
cos θ = v_ox /v_o —–> v_ox =v_o cosθ

ingxenye engu-x (evundlile):
v_ox =v_o cos θ = (10 m/s)(cos 60^o) = (10 m/s)(0.5) = 5 m/s

ingxenye ka-y (eqondile):
v_oy =v_o isono θ = (10 m/s)(isono 60^o) = (10 m/s)(0.5√3) = 5√3 m/s

2. Into ishiya umhlabathi nge-engeli θ = 30^o ngengxenye engu-y yejubane engu-10 m/s. Bala ijubane lokuqala!
Kwaziwa:
I-engeli (θ) = 30^o
ingxenye ka-y (v_oy) = 10 m/s
Kufunwa: Ijubane lokuqala (v)_o)
Isixazululo:
v_oy =v_o isono θ
10 = (v_o)(isono 30^o)
10 = (v_o)(0.5)
v_o = 10/0.5
v_o = 20m/s

3. Ingxenye evundlile yejubane lokuqala ingu-30 m/s kanti ingxenye evundlile yejubane lokuqala ingu-40 m/s. Bala ijubane lokuqala.
Kwaziwa:
Ingxenye evundlile yejubane lokuqala (v)_ox) = 30 m/s
Ingxenye eqondile yejubane lokuqala (v)_oy) = 40 m/s
Kufunwa: Ijubane lokuqala (v)_o)
Isixazululo:
v_o² =v_ox² +v_oy² = 30² 40 +² = 900 + 1600 = 2500
v_o = √2500
v_o = 50m/s

4. Ibhola elincane liphakanyiswa ngokuvundlile ngejubane lokuqala v_o = 6m/s. Bala ingxenye ka-x kanye nengxenye ka-y yejubane lokuqala.
Kwaziwa:
Ijubane lokuqala (v)_o) = 6 m/s
Kufunwa: i-vox kanye ne-v_oy
Isixazululo:
Ibhola lihamba ngokuvundlile ukuze ingxenye yejubane evundlile (v)_ox) = ijubane lokuqala (v_o) = 6 m/s. Ingxenye eqondile yejubane (v)_oy= 0.

[wpdm_package id='545′]

[wpdm_package id='536′]

1. Xazulula ijubane lokuqala libe izingxenye ezivundlile neziqondile
2. Nquma ukufuduka okuvundlile
3. Nquma ukuphakama okuphezulu
4. Nquma isikhathi esinqunyiwe
5. Nquma indawo yezinto
6. Nquma ijubane lokugcina

Funda kabanzi