Mathata a Rarollotsoeng Fisiks ea Motheo

1. Bolo ea maoto e rahiloeng e tloha fatše ka sekhutlo sa θ = 30 ^o ho ea ho sephara ka lebelo la pele la 14 m/s. Bala lebelo la ho qetela pele bolo e otla fatše.

Tse tsejoang:

Sekhutlo ( θ ) = 3 0 ^o

Lebelo la pele (vo _o ) = 14 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Lebelo la ho qetela pele bolo e otla fatše

Tharollo:

Karolo e otlolohileng ea lebelo la pele:

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

Karolo e otlolohileng ea lebelo la pele:

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

Lebelo la ho qetela lehlakoreng le otlolohileng

Khetha tsela e yang hodimo e le e ntle le e theohelang e le e mpe.

Tse tsejoang:

Lebelo la pele (v _o ) = 7 m/s (le letle ho ea holimo)

Ho potlaka ha matla a khoheli (g) = – 10 m/s ² (e mpe ho ea tlase)

Bolelele (h) = 0 (ntho e kgutlelang boemong ba yona ba pele)

Ho batloa: Lebelo la ho qetela (v _t )

Tharollo:

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

v _t = √49 = 7 m/s

Lebelo la ho qetela lehlakoreng le otlolohileng

Lebelo la pele lehlakoreng le otlolohileng ke 7 √ 3 m/s. Lebelo le tsitsitse e le hore lebelo la ho qetela le tšoane le lebelo la pele.

Lebelo la ho qetela pele ntho e fihla fatshe

2. 'Mele o phahamisetsoa holimo ka sekhutlo sa 30 ^o ka bophahamo ba limithara tse 5 ho tloha mohahong. Lebelo la oona la pele ke 10 m/s. Bala lebelo la ho qetela pele ntho e otla fatše! Ho potlaka ha matla a khoheli ke 10 m/s ².

Tse tsejoang:

Sekhutlo ( θ ) = 30 ^o

Bophahamo ba pele (h _o ) = limithara tse 5

Lebelo la pele (vo _o ) = 10 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Lebelo la ho qetela

Tharollo:

Karolo e otlolohileng ea lebelo la pele:

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

Karolo e otlolohileng ea lebelo la pele:

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

Lebelo la ho qetela lehlakoreng le otlolohileng

Tse tsejoang:

Lebelo la pele (v _o ) = 5 m/s (le letle ho ea holimo)

Ho potlaka ha matla a khoheli (g) = – 10 m/s ² (e mpe ho ea tlase)

Bophahamo (h) = -5 m ( bofokodi hobane mobu o ka tlase ho bophahamo ba pele)

Ho batloa: Lebelo la ho qetela (v _t )

Tharollo:

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

v _t = √125 m/s

Lebelo la ho qetela lehlakoreng le otlolohileng

Lebelo la ho qetela lehlakoreng le otlolohileng ke 5 √3 m/s.

Lebelo la ho qetela

3. Bolo e nyane e shebiloe ka ho rapama ka lebelo la pele v _o = 8 m/s ho tloha mohahong o bophahamo ba limithara tse 12. Bala lebelo la ho qetela pele bolo e fihla fatše ! Ho potlaka ha matla a khoheli ke 10 m/s ²

Tse tsejoang:

Bophahamo (h) = limithara tse 12

Lebelo la pele (vo _o ) = 8 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Lebelo la ho qetela (v _t )

Tharollo:

Karolo e otlolohileng ea lebelo la pele:

v _ox = v _o = 8 m/s

Karolo e otlolohileng ea lebelo la pele:

v _oy = 0 m/s

Lebelo la ho qetela lehlakoreng le otlolohileng

e baloang ho sebelisoa equation ea motsamao oa ho oa o lokolohileng.

Tse tsejoang:

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Bolelele (h) = 12 m

Ho batloa: Lebelo la ho qetela (v _t )

Tharollo:

v _t ² = 2 gh = 2(10)(12) = 240

v _t = √240 m/s

Lebelo la ho qetela lehlakoreng le otlolohileng

Lebelo la pele lehlakoreng le bataletseng ke 8 m/s. Lebelo le tsitsitse e le hore lebelo la pele le lekane le lebelo la ho qetela. Kahoo lebelo la ho qetela lehlakoreng le bataletseng ke 8 m/s.

Lebelo la ho qetela

[ID ea sephutheloana sa wpdm='534′]

[ID ea sephutheloana sa wpdm='536′]

1. Rarolla lebelo la pele likarolong tse otlolohileng le tse otlolohileng
2. Fumana sebaka se rapameng
3. Fumana bophahamo bo phahameng ka ho fetisisa
4. Fumana nako ea karohano
5. Fumana boemo ba ntho
6. Fumana lebelo la ho qetela

Bala haholoanyane

Mathata a rarollotsoeng motsamaong o qotsitsoeng - fumana boemo ba ntho

1. 'Mele o phahamisetsoa holimo ka sekhutlo sa 60 ^o ho ea ho se bataletseng ka lebelo la pele la 12 m/s. Fumana boemo ba ntho ka mor'a ho tsamaea motsotsoana o le mong! Ho potlaka ha matla a khoheli ke 10 m/s ².

Tse tsejoang:

Sekhutlo ( θ ) = 60 ^o

Lebelo la pele (vo _o ) = 12 m/s

Nako ea nako (t) = motsotsoana o le mong

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Sebaka sa ntho ka mora ho sisinyeha motsotsoana o le mong

Tharollo:

Karolo e otlolohileng ea lebelo la pele:

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

Karolo e otlolohileng ea lebelo la pele:

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

Sebaka sa ntho ka lehlakoreng le otlolohileng:

Tse tsejoang:

Karolo e otlolohileng ea lebelo (v _x ) = 6 m/s

Nako ea nako (t) = motsotsoana o le mong

Ho batloa: sebaka se bataletseng (x)

Tharollo:

Limithara tse 6 / motsotsoana li bolela hore bolo e tsamaea ho fihlela ho limithara tse 6 motsotsoana o mong le o mong. Sebaka sa bolo ka mor'a ho tsamaea motsotsoana o le mong ke limithara tse 6. Kahoo boemo ba bolo ka lehlakoreng le otlolohileng ke limithara tse 6.

Sebaka sa ntho ka lehlakoreng le otlolohileng:

Khetha tsela e yang hodimo e le e ntle le e theohelang e le e mpe.

Tse tsejoang:

Lebelo la pele (v _o ) = 6 √ 3 m/s (ho ea holimo ho ea holimo)

Nako ea nako (t) = motsotsoana o le mong

Ho potlaka ha matla a khoheli (g) = -10 m/s ² (e mpe ho ea tlase)

Ho batloa: bolelele ka mora ho sisinyeha motsotsoana o le mong

Tharollo:

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 metres.

Sebaka sa ntho ka mora ho sisinyeha motsotsoana o le mong:

Ho falla ho otlolohileng (x) = limithara tse 6

Ho falla ho otlolohileng (y) = limithara tse 5.2

[irp]

2. 'Mele o phahamisetsoa holimo ka sekhutlo sa 30 ^o ho ea ho se bataletseng ho tloha mohahong o bophahamo ba limithara tse 20. Lebelo la oona la pele ke 50 m/s. Bala ho falla ho otlolohileng ka mor'a hore 'mele o tsamaee motsotsoana o le mong! Ho potlaka ha matla a khoheli ke 10 m/s ².

Tse tsejoang:

Sekhutlo ( θ ) = 30 ^o

Bophahamo ba pele (h _o ) = limithara tse 20

Lebelo la pele (v _o ) = 50 m/s

Nako ea nako (t) = motsotsoana o le mong

Ho potlaka ha matla a khoheli (g) = 1 0 m/s ²

Ho batloa: Bolelele (h)

Tharollo:

Karolo e otlolohileng ea lebelo la pele:

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

Bophahamo:

Khetha tsela e yang hodimo e le e ntle le e theohelang e le e mpe.

Tse tsejoang:

Lebelo la pele (vo _o ) = 25 m/s (le holimo ho ea holimo)

Nako ea nako (t) = motsotsoana o le mong

Ho potlaka ha matla a khoheli (g) = -1 0 m/s ² (e mpe ho ya tlase)

Ho batloa: Bolelele (h)

Tharollo:

h = vo _o t + 1/2 gt ^{​​​​2} = (25)(1) + 1/2 (-10)(1 ² ) = 25 + (-5)(1) = 25 - 5 = limithara tse 20.

Bophahamo ba 'mele ka mor'a ho tsamaea motsotsoana o le mong ke limithara tse 20 ka holimo ho moo 'mele o hakantsoeng teng kapa limithara tse 40 ka holimo ho mobu.

[irp]

3. Bolo e nyane e shebiloe ka ho rapama ka lebelo la pele v _o = 10 m/s ho tloha mohahong o bophahamo ba limithara tse 10. Bala ho falla ha bolo ka mor'a ho tsamaea motsotsoana o le mong ! Ho potlaka ha matla a khoheli ke 10 m/s ²

Tse tsejoang:

Bophahamo ba pele (h) = limithara tse 10

Lebelo la pele (vo _o ) = 10 m/s

Nako ea nako (t) = motsotsoana o le mong

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Sebaka sa bolo ka mor'a ho tsamaea motsotsoana o le mong!

Tharollo:

Ho falla ho otlolohileng:

Tse tsejoang:

Karolo e otlolohileng ea lebelo (v _x ) = 10 m/s

Nako ea nako (t) = motsotsoana o le mong

Ho batloa: Sebaka sa ntho

Tharollo:

Limithara tse 10 ka motsotsoana li bolela hore ntho e tsamaea ho fihlela ho limithara tse 10 motsotsoana o mong le o mong. Ho falla ka mor'a ho falla motsotsoana o le mong ke limithara tse 10. Kahoo ho falla ka ho rapama ke limithara tse 10.

Ho falla ho otlolohileng:

E baloa e le motsamao oa ho oa o lokolohileng.

Tse tsejoang:

Nako ea nako (t) = motsotsoana o le mong

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Bolelele ka mora ho sisinyeha motsotsoana o le mong (h)

Tharollo:

h = 1/2 gt ^{​​​​2} = 1/2 (10)(1 ² ) = (5)(1) = limithara tse 5.

Kamora motsotsoana o le mong, ntho e oela ho fihlela ho limithara tse 5. Bophahamo ka holim'a mobu = limithara tse 10 - limithara tse 5 = limithara tse 5.

Sebaka sa ntho ka mora ho tsamaya motsotsoana o le mong:

Sebaka sa ntho ka lehlakoreng le otlolohileng (x) = limithara tse 10

Sebaka sa ntho ka lehlakoreng le otlolohileng (y) = limithara tse 5

[irp]

[ID ea sephutheloana sa wpdm='532′]

[ID ea sephutheloana sa wpdm='536′]

1. Rarolla lebelo la pele likarolong tse otlolohileng le tse otlolohileng
2. Fumana sebaka se rapameng
3. Fumana bophahamo bo phahameng ka ho fetisisa
4. Fumana nako ea karohano
5. Fumana boemo ba ntho
6. Fumana lebelo la ho qetela

Bala haholoanyane

Mathata a rarollotsoeng motsamaong oa projectile - fumana nako ea ho arohana

1. Bolo ea maoto e rahiloeng e tloha fatše ka sekhutlo sa θ = 30 ^o ho ea ho sephara ka lebelo la pele la 10 m/s. Bala nako ea ho fihla bophahamong bo phahameng ka ho fetisisa! Ho potlaka ha matla a khoheli ke 10 m/s ².

Tse tsejoang:

Sekhutlo (θ) = 3 0 ^o

Lebelo la pele (vo _o ) = 10 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Nako ea ho fihla bophahamong bo phahameng ka ho fetisisa

Tharollo:

Karolo e otlolohileng ea lebelo la pele:

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

Nako e sa fellang ho fihlela bophahamo bo phahameng ka ho fetisisa e laolwa ke equation ya motsamao o otlolohileng . Khetha tataiso e yang hodimo e le e ntle le e theohelang e le e mpe.

Tse tsejoang:

Lebelo la pele (v _o ) = 5 m/s (le letle ho ea holimo)

Ho potlaka ha matla a khoheli (g) = – 10 m/s ² (e mpe ho ea tlase)

Lebelo la ho qetela bophahamong bo phahameng ka ho fetisisa (v _t ) = 0

Ho batloa: nako e sa fellang (t)

Tharollo:

v _t = v _o + g t

0 = 5 + (-10)t

0 = 5 – 10 t

5 = 10 t

t = 5/10 = 0.5 s

[irp]

2. 'Mele o phahamisetsoa holimo ka sekhutlo sa 30 ^o ho ea ho se bataletseng ka lebelo la pele la 30 m/s. Bala nako ea ho fofa! Potlakiso ea matla a khoheli ke 10 m/s ².

Tse tsejoang:

Sekhutlo (θ) = 3 0 ^o

Lebelo la pele (vo _o ) = 8 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Nako e sa fellang pele setopo se thula fatshe

Tharollo:

Karolo e otlolohileng ea lebelo la pele:

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

Re qala ka ho bala nako e pakeng tsa nako ho fihlela bophahamo bo phahameng ka ho fetisisa re sebelisa equation ea motsamao o otlolohileng.

Khetha tsela e yang hodimo e le e ntle le e theohelang e le e mpe.

Tse tsejoang:

Lebelo la pele (v _o ) = 4 m/s (le holimo ho ea holimo)

Ho potlaka ha matla a khoheli (g) = – 10 m/s ² (e mpe ho ea tlase)

Lebelo la ho qetela bophahamong bo phahameng ka ho fetisisa (v _t ) = 0

Ho batloa: Nako e sa fellang (t)

Tharollo:

v _t = v _o + g t

0 = 4 + (-10)t

0 = 4 – 10 t

4 = 10 t

t = 4/10 = 0,4 s

Nako ea ho fihla bophahamong bo phahameng ka ho fetisisa ke 0.4 s.

Nako moeeng ke 2 x 0.4 s = 0.8 s.

[irp]

3. 'Mele o phahamisetsoa holimo ka sekhutlo sa 30 ^o ka bophahamo ba limithara tse 10 ho tloha mohahong. Lebelo la oona la pele ke 40 m/s. 'Mele o nka nako e kae ho fihla fatše? Ho potlaka ha matla a khoheli ke 10 m/s ².

Tse tsejoang:

Sekhutlo (θ) = 3 0 ^o

Bophahamo ba pele (h _o ) = limithara tse 10

Lebelo la pele (vo _o ) = 40 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Nako moeeng (t)

Tharollo:

Karolo e otlolohileng ea lebelo la pele:

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

Re qala ka ho bala nako e pakeng tsa nako ho fihlela bophahamo bo phahameng ka ho fetisisa re sebelisa equation ea motsamao o otlolohileng.

Khetha tsela e yang hodimo e le e ntle le e theohelang e le e mpe.

Tse tsejoang:

Lebelo la pele (v _o ) = 20 m/s (le holimo ho ea holimo)

Ho potlaka ha matla a khoheli (g) = – 10 m/s ² (e mpe ho ea tlase)

Lebelo la ho qetela sehlohlolong (v _t ) = 0

Ho batloa: Nako e sa fellang (t)

Tharollo:

v _t = v _o + g t

0 = 20 + (-10)t

0 = 20 – 10 t

20 = 10 t

t = 20/10 = metsotsoana e 2

Nako moeeng = 2 x 2 metsotsoana = 4 metsotsoana.

Ntho e hodimo ho dimithara tse 10 ka hodima lefatshe. Metsotswana e 4 ke nako ya ho fihla sebakeng se bapileng le boemo ba pele. Bolo e ntse e theohela tlase.

Nako ea ho fihla fatše e baloa ho sebelisoa equation ea motsamao oa ho oa ka bolokolohi.

Tse tsejoang:

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Bophahamo (h) = limithara tse 10

Ho batloa: Nako e sa fellang (t)

Tharollo:

h = 1/2 gt ^{​​​​2}

10 = 1/2 (10) t ²

10 = 5 t ²

t ² = 10/5 = 2

t = √2 = metsotsoana e 1.4

Nako e sa fellang = metsotsoana e 1.4.

Nako yohle e kgaotsweng = metsotsoana e 4 + metsotsoana e 1.4 = metsotsoana e 5.4.

[irp]

4. Bolo e nyane e shebiloe ka ho otloloha ka lebelo la pele v _o = 15 m/s ho tloha mohahong o bophahamo ba limithara tse 5. Bala nako moeeng ! Ho potlaka ha matla a khoheli ke 10 m/s ²

Tse tsejoang:

Bophahamo (h) = limithara tse 5

Lebelo la pele (vo _o ) = 15 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Nako moeeng (t)

Tharollo:

Nako moeeng e baloa ho sebelisoa equation ea motsamao o oang ka bolokolohi.

Tse tsejoang:

Bophahamo (h) = limithara tse 5

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Nako e sa fellang (t)

Tharollo:

h = 1/2 gt ^{​​​​2}

5 = 1/2 (10) t ²

5 = 5 t ²

t ² = 5/5 = 1

t = √1 = motsotsoana o le mong

[irp]

[ID ea sephutheloana sa wpdm='531′]

[ID ea sephutheloana sa wpdm='536′]

1. Rarolla lebelo la pele likarolong tse otlolohileng le tse otlolohileng
2. Fumana sebaka se rapameng
3. Fumana bophahamo bo phahameng ka ho fetisisa
4. Fumana nako ea karohano
5. Fumana boemo ba lintho
6. Fumana lebelo la ho qetela

Bala haholoanyane

Mathata a rarollotsoeng motsamaong oa projectile - fumana bophahamo bo phahameng ka ho fetisisa

1. Bolo ea maoto e rahiloeng e tloha fatše ka sekhutlo sa θ = 60 ^o 'me lebelo la pele la ho rapama ke 10 m/s. Bala bophahamo bo phahameng ka ho fetisisa! Ho potlaka ha matla a khoheli ke 10 m/s ².

Tse tsejoang:

Sekhutlo (θ) = 60 ^o

Lebelo la pele (vo _o ) = 10 m/s

Ho batloa: Bolelele bo phahameng ka ho fetisisa (h)

Tharollo:

Karolo e otlolohileng ea lebelo la pele:

sebe 60 ^o = v _oy / v _o

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

Khetha tsela e yang hodimo e le e ntle le e theohelang e le e mpe.

Tse tsejoang:

Ho potlaka ha matla a khoheli (g) = -10 m/s ² (e mpe ho ea tlase)

Karolo e otlolohileng ea lebelo la pele (v _oy ) = + 5 √ 3 m/s (e ntle ho ea holimo)

Lebelo la ho qetela bophahamong bo phahameng ka ho fetisisa (v _ty ) = 0

Ho batloa: Bolelele bo phahameng ka ho fetisisa (h)

Tharollo:

v _t ² = v _o ² + 2 gh

0 ² = ( 5 √ 3) ² + 2 (-10) h

0 = 2 5 ( 3) – lihora tse 20

0 = 75 – 20 dihora

75 = lihora tse 20

h = 75 / 20

h = limithara tse 3.75

Bophahamo bo phahameng ka ho fetisisa ke limithara tse 3.75.

[irp]

2. 'Mele o phahamisetsoa holimo ka sekhutlo sa 30 ^{o, 'me bophahamo ba oona bo rapameng ho tloha mohahong bo fihla ho limithara tse 20. Lebelo la oona la pele ke 4 m/s. Bala bophahamo bo phahameng ka ho fetisisa! Ho potlaka ha matla a khoheli ke 10} m/s.

Tse tsejoang:

Sekhutlo (θ) = 30 ^o

Bophahamo ba pele (h) = limithara tse 20

Lebelo la pele (vo _o ) = 4 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Bophahamo bo phahameng ka ho fetisisa (h)

Tharollo:

Karolo e otlolohileng ea lebelo la pele:

sebe 30 ^o = v _oy / v _o

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

Khetha tsela e yang hodimo e le e ntle le e theohelang e le e mpe.

Tse tsejoang:

Ho potlaka ha matla a khoheli (g) = -10 m/s ² (e mpe ho ea tlase)

Karolo e otlolohileng ea lebelo la pele (v _oy ) = +2 m/s (e ntle ho ea holimo)

Lebelo la ho qetela bophahamong bo phahameng ka ho fetisisa (v _ty ) = 0

Ho batloa: Bolelele bo phahameng ka ho fetisisa

Tharollo:

Bophahamo bo phahameng ka ho fetisisa:

v _t ² = v _o ² + 2 gh

0 ² = 2 ² + 2 (-10) h

0 = 4 – 20 dihora

4 = lihora tse 20

h = 4 / 20

h = limithara tse 0.2

Bophahamo bo phahameng ka ho fetisisa ke limithara tse 0.2 + limithara tse 20 = limithara tse 20.2.

[irp]

[ID ea sephutheloana sa wpdm='528′]

[ID ea sephutheloana sa wpdm='536′]

1. Rarolla lebelo la pele likarolong tse otlolohileng le tse otlolohileng
2. Fumana sebaka se rapameng
3. Fumana bophahamo bo phahameng ka ho fetisisa
4. Fumana nako ea karohano
5. Fumana boemo ba lintho
6. Fumana lebelo la ho qetela

Bala haholoanyane

Mathata a rarollotsoeng motsamaong oa projectile - fumana sebaka se otlolohileng

1. Bolo ea maoto e rahiloeng e tloha fatše ka sekhutlo sa θ = 60 ^o 'me lebelo la pele la 16 m/s le tla nka nako e kae pele bolo e otla fatše?

Tse tsejoang:

Sekhutlo ( θ ) = 60 ^o

Lebelo la pele (vo _o ) = 16 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Ho falla ho otlolohileng (x)

Tharollo:

Karolo e otlolohileng ea lebelo la pele:

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

Karolo e otlolohileng ea lebelo la pele:

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

Motsamao o rotohileng o ka utlwisiswa ka ho sekaseka dikarolo tse otlolohileng le tse otlolohileng tsa motsamao ka thoko. Motsamao wa x o etsahala ka lebelo le sa fetoheng mme motsamao wa y o etsahala ka ho potlaka ha matla a khoheli kamehla.

Nako moeeng

Nako eo e dulang moeeng e laolwa ke motsamao wa y. Re qala ka ho fumana nako re sebedisa motsamao wa y ebe re sebedisa boleng bona ba nako ho di-equation tsa x ( equation ya lebelo le sa fetoheng ).

Khetha tsela e yang hodimo e le e ntle le e theohelang e le e mpe.

Tse tsejoang:

Lebelo la pele (v _o ) = 8 √ 3 m/s (v _o ho ya hodimo)

Ho potlaka ha matla a khoheli (g) = -10 m/s ² (g ho ea tlase)

Bolelele (h) = 0 (bolo e kgutletse boemong bo le bong)

Ho Batloa: Nako Moeeng

Tharollo:

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 = metsotsoana e 2.8

Ho falla ho otlolohileng

Tse tsejoang:

Lebelo (v) = 8 m/s

Nako ea ho arohana (t) = metsotsoana e 2.8

Ho Batloa: Ho falla

Tharollo:

x = vt = (8 m/s)(2.8 s) = limithara tse 22.4

Sebaka se otlolohileng sa ho falla ke limithara tse 22.4.

[irp]

2. 'Mele o phahamisetsoa holimo ka sekhutlo sa 60 ^{o, 'me bophahamo ba oona bo tloha mohahong bo fihla ho limithara tse 50. Lebelo la oona la pele ke 30 m/s. Bala ho falla ho rapameng! Ho potlaka ha matla a khoheli ke 10} m/s.

Tse tsejoang:

Sekhutlo ( θ ) = 60 ^o

Bolelele (h) = 15 m

Lebelo la pele (v _o ) = 30 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: x

Tharollo:

Karolo e otlolohileng ea lebelo la pele ::

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

Karolo e otlolohileng ea lebelo la pele:

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

Nako moeeng

Re qala ka ho fumana nako re sebelisa motsamao oa y ebe re sebelisa boleng bona ba nako ho li-equation tsa x (equation ea lebelo le sa fetoheng). Khetha ho ea holimo e le ntle le ho ea tlase e le mpe.

Tse tsejoang:

Lebelo la pele (v _o ) = 15 √ 3 m/s (ho ea holimo ho ea holimo)

Ho potlaka ha matla a khoheli (g) = -10 m/s ² (e mpe ho ea tlase)

Holimo (h) = -50 (Botlaaseng ba limithara tse 50 ka tlase ho sebaka sa pele)

Ho batloa: t

Tharollo:

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 nako u sebelisa foromo ena:

a = 5, b = – 15 √ 3, c = – 50

Nako moeeng ke metsotsoana e 6.7.

Ho falla ho otlolohileng:

Tse tsejoang:

Lebelo (v) = 15 m/s

Nako ea ho arohana (t) = metsotsoana e 6.7

Ho batloa: ho falla

Tharollo:

s = vt = (15 m/s)(6.7 s) = limithara tse 100.5

Sebaka se otlolohileng sa ho falla ke limithara tse 100.5.

[irp]

3. Bolo e nyane e shebiloe ka ho rapama ka lebelo la pele v _o = 10 m/s ho tloha mohahong o bophahamo ba limithara tse 10. Bala ho falla ho rapameng ! Ho potlaka ha matla a khoheli ke 10 m/s ²

Tse tsejoang:

Bolelele (h) = 10 m

Lebelo la pele (vo _o ) = 10 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: x

Tharollo:

Karolo e otlolohileng ea lebelo la pele = lebelo la pele = 10 m/s.

Nako moeeng

Nako moeeng e baloa ho sebelisoa equation ea motsamao oa ho oa mahala.

Tse tsejoang:

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Bophahamo (h) = limithara tse 10

Ho batloa: t

Tharollo:

h = 1/2 gt ^{​​​​2}

10 = 1/2 (10) t ²

10 = 5 t ²

t ² = 10 / 5 = 2

t = √2 = metsotsoana e 1.4

Ho falla ho otlolohileng

Ho falla ho rapameng ho balwa ho sebediswa equation ya motsamao ka lebelo le sa fetoheng.

Tse tsejoang:

Lebelo (v) = 10 m/s

Nako ea ho arohana (t) = metsotsoana e 1.4

Ho batloa: x

Tharollo:

s = vt = (10 m/s)(1.4 s) = limithara tse 14

Sebaka se otlolohileng sa ho falla ke limithara tse 14.

[irp]

[ID ea sephutheloana sa wpdm='526′]

[ID ea sephutheloana sa wpdm='536′]

1. Rarolla lebelo la pele likarolong tse otlolohileng le tse otlolohileng
2. Fumana sebaka se rapameng
3. Fumana bophahamo bo phahameng ka ho fetisisa
4. Fumana nako ea karohano
5. Fumana boemo ba lintho
6. Fumana lebelo la ho qetela

Bala haholoanyane

Mathata a rarollotsoeng motsamaong oa projectile - a rarollotse lebelo la pele likarolong tse otlolohileng le tse otlolohileng

1. Bolo ea maoto e rahiloeng e tloha fatše ka sekhutlo θ = 60^o ka lebelo la 10 m/s. Bala likarolo tsa lebelo la pele!
Tse tsejoang:
Sekhutlo (θ) = 60^o
Lebelo la pele (v)_o) = 10 m/s
Ho batloa: v_ox le v_oy
Tharollo:
Rarolla lebelo la pele ho karolo ea x (e otlolohileng) le karolo ea y (e otlolohileng).
sebe θ = v_oy /v_o —–> v_oy = v_o sebe θ
cos θ = v_ox /v_o —–> v_ox = v_o cos θ

karolo ea x (e otlolohileng):
v_ox = v_o cos θ = (10 m/s)(cos 60^o) = (10 m/s)(0.5) = 5 m/s

karolo ea y (e otlolohileng):
v_oy = v_o sebe θ = (10 m/s)(sebe 60^o) = (10 m/s)(0.5√3) = 5√3 m/s

[irp]

2. Ntho e tloha fatshe ka sekhutlo θ = 30^o ka karolo ea y ea lebelo la 10 m/s. Bala lebelo la pele!
Tse tsejoang:
Sekhutlo (θ) = 30^o
karolo ea y (v_oy) = 10 m/s
Ho batloa: Lebelo la pele (v)_o)
Tharollo:
v_oy = v_o sebe θ
10 = (v_o)(sebe 30^o)
10 = (v_o)(0.5)
v_o = 10/0.5
v_o = 20 m/s

[irp]

3. Karolo e otlolohileng ea lebelo la pele ke 30 m/s mme karolo e otlolohileng ea lebelo la pele ke 40 m/s. Bala lebelo la pele.
Tse tsejoang:
Karolo e otlolohileng ea lebelo la pele (v)_ox) = 30 m/s
Karolo e otlolohileng ea lebelo la pele (v)_oy) = 40 m/s
Ho batloa: Lebelo la pele (v)_o)
Tharollo:
v_o² = v_ox² + v_oy² = 30² + 40² 900 + 1600 = 2500
v_o = √2500
v_o = 50 m/s

[irp]

4. Bolo e nyane e shebiloe ka ho otloloha ka lebelo la pele v_o = 6m/s. Bala karolo ea x le karolo ea y ea lebelo la pele.
Tse tsejoang:
Lebelo la pele (v)_o) = 6 m/s
Ho batloa: vox le v_oy
Tharollo:
Bolo e tsamaea ka ho rapama e le hore karolo e bataletseng ea lebelo (v)_ox) = lebelo la pele (v_o) = 6 m/s. Karolo e otlolohileng ea lebelo (v_oy= 0.

[irp]

[ID ea sephutheloana sa wpdm='545′]

[ID ea sephutheloana sa wpdm='536′]

1. Rarolla lebelo la pele likarolong tse otlolohileng le tse otlolohileng
2. Fumana sebaka se rapameng
3. Fumana bophahamo bo phahameng ka ho fetisisa
4. Fumana nako ea karohano
5. Fumana boemo ba lintho
6. Fumana lebelo la ho qetela

Bala haholoanyane