projectile-motion

Isihloko mayelana nokunyakaza kwe-Projectile nezinkinga zesampula ngezixazululo

Ijubane lokuqala (v)_o) kanye nengxenye yejubane lokuqala (v_ox kanye v_oy)

Into ekunyakaza kwayo okufana nokufana nokufana okuhlala kunesivinini sokuqala. Ngenxa yokuthi ukunyakaza okufana nokufana kuyinhlanganisela yokunyakaza eziqondisweni ezivundlile neziqondile, ijubane lokuqala liphinde libe nezingxenye ezivundlile neziqondile.

Uma into ihamba ngendlela eyisimangaliso njengaseZithombeni 1 no-3 khona-ke ijubane lokuqala ohlangothini oluvundlile (v _ox ) kanye nejubane lokuqala ohlangothini oluvundlile (v _oy ) kubalwa kusetshenziswa i-equation:

v _ox = vo _o cos θ

v _oy = vo _o sin θ

Uma into ihamba ngendlela eyisimangaliso njengomfanekiso 2 khona-ke v _o = v _ox (v _oy = 0)

Ijubane (v)_x kanye v_y) kanye nesikhundla (x kanye no-y)

Ijubane eliqondiswe ngokuqondile nangokuqondile ngesikhathi esithile libalwa kusetshenziswa i-equation:

v _x = v _ox = okungaguquki ( ukunyakaza okuqondile okufanayo )

v _y = v _oy + gt noma v _y ² = v _oy ² + 2 gh (ukunyakaza kokuwa ngokukhululeka)

Indawo yezinto eziqondisweni ezivundlile (x) kanye neziqondile (y) ngesikhathi esithile ibalwa kusetshenziswa i-equation:

x = v _ox t

y = v _oy t + 1⁄2 gt ²

Umphumela wejubane (v) kanye nesikhundla (h)

Ijubane eliphumela ngesikhathi esithile libalwa kusetshenziswa i-equation:

Isiqondiso sezinto ngesikhathi esithile sibalwa kusetshenziswa i-equation:

amanothi:

1. Ingxenye evundlile yokunyakaza kwe-parabolic ibhekwa njengokunyakaza okuqondile okufanayo, ngakho-ke i-vox = vx ihlala ingaguquguquki

2. Ingxenye eqondile yokunyakaza kwe-parabolic ibhekwa njengokunyakaza kokuwa okukhululekile, ngakho-ke uma into ihamba nge-parabolic, njengeZithombe 1 no-3, ingxenye eqondile yejubane lento ekuphakameni okuphezulu ingu-zero (v _y = 0). Uma uphonsa imabula iqonde phezulu ekuphakameni okuphezulu, into iphumula isikhashana (v _y = 0) ngaphambi kokujika iye phansi. Ngakho-ke, ijubane lento ehambisa i-parabolic ekuphakameni okuphezulu = v _x = v _ox

3. Uma into ihamba ngendlela ephambene njengoba kuboniswe kuMfanekiso 2, ingxenye eqondile yokunyakaza ngendlela ephambene ibhekwa njengokunyakaza kokuwa okungenantambo. Uma into ihamba ngendlela ephambene njengeMifanekiso 1 no-3 khona-ke ingxenye eqondile yokunyakaza ngendlela ephambene ibhekwa njengokunyakaza okuqonde phezulu).

Izinkinga zesampula:

1. Inhlamvu idutshulwa ngendlela evundlile ngesivinini sokuqala esingu-20 m/s. Uma isibhamu singamamitha ama-5 ngaphezu komhlabathi, thola:

(a) isikhathi emoyeni

(b) ukuphakama okuphezulu

(c) ibanga elivundlile

(d) isivinini senhlamvu uma ishaya phansi

Isixazululo:

Ukunyakaza ohlangothini oluvundlile kuhlaziywa njengokunyakaza okuqondile okufanayo, kuyilapho ukunyakaza ohlangothini oluqondile kuhlaziywa njengokunyakaza kokuwa okukhululekile.

Kwaziwa:

i-v _ox = 20 m/s, i-v _oy = 0 m/s, i-h = 5 m, i-g = 9.8 m/s ²

a) Isikhathi emoyeni

Isixazululo sinjengokunquma isikhawu sesikhathi (t) ekunyakazeni kokuwa okukhululekile.

Kwaziwa: v _oy = 0 m/s, h = 5 m, g = 9.8 m/s ²

Kufunwa: t

b) Ukuphakama okuphezulu

Ukuphakama okuphezulu = h = amamitha angu-5.

c) Ibanga elivundlile (d)

Isixazululo sifana nokunquma ibanga ekuhambeni okuqondile okufanayo

Kwaziwa: v _ox = 20 m/s, t = 1 umzuzwana

Kufunwa: d

d = vt

d = (20 m/s)(1 s) = 20 m

d) Shesha lapho inhlamvu ishaya phansi

v _tx = v _ox = 20 m/s

v _ty = ?

Okokuqala, sibala isivinini sokugcina ohlangothini oluqondile (vty). Isixazululo sifana nokunquma isivinini sokugcina sokunyakaza kwe-free-fall.

Kwaziwa: v _oy = 0, g = 9.8 m/s ² , t = 1 s

Kufunwa: v _ty

v _t = vo _o + gt —> vo = 0

v _t = gt

v _t = (9.8 m/s ² )(1 s)

v _t = 9.8 m/s

Isivinini senhlamvu uma ishaya phansi:

Isiqondiso senhlamvu:

Ngoba i-v _tx isendleleni ye-x-axis elungile (ngakwesokudla) kanti i-v _ty isendleleni ye-y-axis engalungile (phansi),

isiqondiso senhlamvu uma ishaya phansi singu -26.1 ^o kuya ku-x-axis enhle (bheka isithombe esingezansi).

2. Isibhamu sadubula inhlamvu ngesivinini esingama-30 ^o siye endaweni evundlile ngesivinini esingama-60 m/s. Thola:

(a) ukuphakama okuphezulu

(b) isivinini senhlamvu ekuphakameni okuphezulu

(c) isikhathi emoyeni

(d) ibanga elivundlile

(e) isivinini senhlamvu uma ishaya phansi. Ake sithi umhlabathi uyisicaba. 🙂

Isixazululo:

Ukunyakaza ohlangothini oluvundlile kuhlaziywa njengokunyakaza okuqondile okufanayo, ukunyakaza ohlangothini oluqondile kuhlaziywa njengokunyakaza okuqondile okuphezulu.

Kwaziwa: v _o = 60 m/s, theta = 30 ^o.

Ngokusekelwe kudatha eyaziwayo, siqala ngokubala izingxenye eziqondile (v _oy ) kanye nezivundlile (v _ox ) zejubane lokuqala (v _o ).

a) Ukuphakama okuphezulu (h)

Isixazululo sifana nokunquma ukuphakama okuphezulu ekunyakazeni okuqonde phezulu.

Kwaziwa:

v _inkabi = v _o cos θ = (60)(cos 30) = (60)(0.87) = 52 m/s

v _oy = v _o isono θ = (60)(isono 30) = (60)(0.5) = 30 m/s

a) Ukuphakama okuphezulu (h)

Isixazululo sifana nokunquma ukuphakama okuphezulu ekunyakazeni okuqonde phezulu.

Kwaziwa:

v _oy = 30 m/s (lesi yisivinini sokuqala senhlamvu)

v _ty = 0 m/s (Ekuphakameni okuphezulu, ijubane eliqondile lenhlamvu = 0 m/s. Lesi yisivinini sokugcina.)

g = – 9.8 m/s ²

Kufunwa: h

v _t ² = v _o ² + 2 gh

0 ² = 30 ² + 2 (-9.8) h

0 = 900 – 19.6 amahora

900 = amahora angu-19.6

h = 900/19.6

h = amamitha angu-45.9

Ukuphakama okuphezulu okufinyelelwe yinhlamvu = amamitha angu-45.9.

b) Isivinini sokuphakama okuphezulu

Ekuphakameni okuphezulu, isivinini esiqonde ngqo = 0 m/s. Ekuphakameni okuphezulu, kukhona isivinini kuphela esiqonde ngqo. Isivinini esiqonde ngqo esiqonde ngqo silingana nesivinini sokuqala esiqonde ngqo, esingama-52.2 m/s. Isiqondiso sejubane esiqonde ngqo sihlala singaguquguquki, okungukuthi, esiqondisweni se-x-axis enhle (uma ukunyakaza kwento kuchazwe kumdwebo ongenhla)

c) Isikhathi emoyeni

Isixazululo sifana nokunquma isikhawu sesikhathi (t) engxoxweni yokunyakaza okuqonde phezulu.

Kwaziwa:

v _oy = 30 m/s (lesi isivinini sokuqala senhlamvu ohlangothini oluqondile)

g = – 9.8 m/s ²

h = 0 m (lapho inhlamvu ibuyela phansi, ukufuduka kwenhlamvu ohlangothini oluqondile = 0 m)

Kufunwa: t

h = vot ₊ ½ gt ²

0 = (30) t + ½ (-9.8 m/s ² ) t ²

0 = (30) t – 4.9 t ²

(30) t = 4.9 t ²

30 = 4.9 t

t = 30 / 4.9

t = imizuzwana emi-6.12

Isikhathi emoyeni = imizuzwana engu-6.12

d) Ibanga elivundlile (d)

Isixazululo sifana nokunquma ibanga (d) ekunyakazeni okuqondile okufanayo.

Kwaziwa:

v _ox = 52.2 m/s

t = imizuzwana emi-6.12

Kufunwa: d

d = vt = (52.2 m/s)(6.12 imizuzwana) = 319.5 m

e) Shesha lapho inhlamvu ishaya phansi

v _tx = v _ox = 52.2 m/s

v _ty = ?

Okokuqala, sibala ijubane lokugcina ohlangothini oluqondile (v _ty ). Isixazululo ukuthola ijubane lokugcina ekunyakazeni okuqonde phezulu okuqondile.

Okufunwayo: v _oy = 30 m/s, g = -9.8 m/s ² , t = 6.12 imizuzwana

Kufunwa: v _ty

v _ty = v _oy + gt

v _ty = (30) + (-9.8)(6.12)

v _ty = (30) – (60)

v _ty = -30 m/s

Uphawu olunegethivu lubonisa ukuthi isiqondiso sejubane lokugcina sehla. Qaphela ukuthi ijubane lokuqala ohlangothini oluqondile lilingana nejubane lokugcina ohlangothini oluqondile.

Isivinini senhlamvu uma ishaya phansi:

Isiqondiso senhlamvu:

Njengoba i-v _tx isendleleni ye-x-axis elungile (ngakwesokudla) kanti i-vty isendleleni ye-y-axis engalungile (phansi),

isiqondiso sejubane lenhlamvu uma ishaya phansi singu -30 ^o cishe nge-x-axis enhle (bheka isithombe esingezansi).

3. Ibhola liphonswa lisuka emaphethelweni esakhiwo esingamamitha angu-50 ukuphakama ngesivinini sokuqala esingu-10 m/s. Uma ibhola liphonswa ku-30o cishe ngokuvundlile, thola:

(a) isikhathi lapho ibhola lifika phansi

(b) isivinini sebhola uma lishaya phansi

(c) ibanga eliqondile elingafinyelelwa yibhola lilinganiswa kusukela emaphethelweni esakhiwo

(d) ukuphakama okuphezulu okufinyelelwe yibhola

Isixazululo:

Okokuqala, sibala ingxenye eqondile (v _oy ) kanye nengxenye evundlile (v _ox ) yejubane lokuqala (v _o ).

v _ox = vo _o cos 30 ^o = (10 m/s)(0.87) = 8.7 m/s

v _oy = vo _o sin 30 ^o = (10 m/s)(0.5) = 5 m/s

a) Isikhathi esiphakathi ibhola lifika phansi

Isixazululo sifana nokunquma isikhawu sesikhathi (t) ngokunyakaza okuqonde phezulu. Ubukhulu bevektha eqondiswe phezulu bukhethwa ukuthi bube buhle, ubukhulu bevektha eqondiswe phansi bukhethwa ukuthi bube buhle. Indawo yebhola lapho iphonswa khona ikhethwa njengendawo yokubhekisela. u-h uhle ngoba ubuso bomhlaba bungaphansi kwendawo yokubhekisela, u-g uhle ngoba indlela yokusheshisa amandla adonsela phansi iphansi.

Kwaziwa:

v _oy = 5 m/s, h = – 5 m, g = – 9.8 m/s ²

Kufunwa: t

h = vot ₊ ½ gt ²

-5 = 5 t + ½ (-9.8) t ²

-5 = 5 t – 4.9 t ²

-4.9 t ² + 5 t + 5 = 0

Sebenzisa ifomula ye-quadratic:

Isikhathi emoyeni = isikhathi esiphakathi selokhu ibhola laphonswa ukuze lifike phansi = imizuzwana eyi-1.64.

b) Isivinini sebhola uma lishaya ubuso bomhlaba

v _tx = v _ox = v _x = 8.7 m/s

v _ty = ?

Okokuqala, sibala isivinini sokugcina ohlangothini oluqondile (v _ty ). Isixazululo sifana nokunquma isivinini sokugcina ekunyakazeni okuqonde phezulu okuqondile.

Kwaziwa: v _oy = 5 m/s, g = -9.8 m/s ² , t = 1.64 imizuzwana

Kufunwa: v _ty

v _ty = v _oy + gt

v _ty = 5 + (-9.8)(1.64)

v _ty = 5 – 16

v _ty = -11 m/s

Uphawu olubi lubonisa ukuthi isiqondiso sejubane lokugcina sehla.

Isivinini senhlamvu uma ishaya phansi:

Isiqondiso sejubane lenhlamvu = isiqondiso sokunyakaza kwenhlamvu uma ishaya phansi:

Isivinini senhlamvu uma ishaya phansi:

Ngoba i-vtx isendleleni ye-x-axis elungile (ngakwesokudla) kanti i-vty isendleleni ye-y-axis engalungile (phansi),

isiqondiso senhlamvu uma ishaya phansi singu--52o cishe nge-x-axis enhle (bheka isithombe esingezansi).

c) Ibanga elivundlile elingafinyelelwa yibhola lilinganiswa kusukela emaphethelweni esakhiwo

Isixazululo sifana nokunquma ibanga elihanjiwe (d) ekuhambeni okuqondile okufanayo.

Kwaziwa: t = 1.64 imizuzwana, v _x = 8.7 m/s

Kufunwa: d

d = vt = (8.7 m/s)(1.64 s) = 14.3 m

d) Ukuphakama okuphezulu okufinyelelwe yibhola

Okuyaziwayo: v _oy = 5 m/s, v _ty = 0 m/s ( ingxenye eqondile yejubane ekuphakameni okuphezulu = 0 m/s), g = -9.8 m/s ².

Kufunwa: h

v _ty ² = v _oy ² + 2 gh

0 m/s = (5 m/s) ² + 2(-9.8 m/s ² )(h)

0 m/s = 25 (m/s) ² + (-19.6 m/s ² )(h)

25 (m/s) ² = -19.6 m/s 2 (h)

h = 25 (m/s) ² : -19.6 m/s ² = 1.3 amamitha

Ukuphakama okuphezulu okufinyelelwa yibhola = amamitha ayi-1.3 ngaphezu kwesakhiwo = 1.3 m + 50 m = amamitha ayi-51.3 ngaphezu komhlabathi.