projectile-motsi

Labari game da motsin Projectile da samfuran matsalolin mafita

Saurin farko (v)_o) da kuma ɓangaren saurin farko (v_ox kuma v_oy)

Abu wanda motsin parabolic ɗinsa koyaushe yana da saurin farko. Saboda motsin parabolic haɗuwa ne na motsi a cikin kwatancen kwance da tsaye, saurin farko shima yana da abubuwan da ke kwance da tsaye.

Idan abin yana motsawa ta hanyar parabolical kamar yadda yake a cikin Figures 1 da 3 to ana ƙididdige saurin farko a cikin alkiblar kwance (v _ox ) da saurin farko a cikin alkiblar tsaye (v _oy ) ta amfani da lissafi:

v _ox = v _o cos θ

v _oy = v _o sin θ

Idan abin da ke motsi a matsayin siffa ta 2 to v _o = v _ox (v _oy = 0)

Gudu (v_x kuma v_y) da Matsayi (x da y)

Ana ƙididdige saurin da ke cikin kwatancen kwance da tsaye a wani takamaiman lokaci ta amfani da lissafi:

v _x = v _ox = akai-akai ( motsi mai layi ɗaya )

v _y = v _oy + gt ko v _y ² = v _oy ² + 2 gh (motsin faɗuwa kyauta)

Ana ƙididdige matsayin abubuwa a cikin kwatancen kwance (x) da tsaye (y) a wani takamaiman lokaci ta amfani da lissafin:

x = v _ox t

y = v _oy t + 1⁄2 gt ²

Sakamakon gudu (v) da matsayi (h)

Ana ƙididdige saurin sakamako a tazara ta lokaci ta amfani da lissafi:

Ana ƙididdige alkiblar abubuwa a wani takamaiman lokaci ta amfani da lissafin:

Notes:

1. Ana ɗaukar ɓangaren kwance na motsi na parabolic a matsayin motsi na layi ɗaya, don haka vox = vx koyaushe yana da daidaito

2. Ana kallon ɓangaren tsaye na motsi na parabolic a matsayin motsi na faɗuwa kyauta, don haka idan abin ya motsa parabolic, kamar Hotuna na 1 da 3, ɓangaren tsaye na saurin abu a matsakaicin tsayi shine sifili (v _y = 0). Idan ka jefa marmara a tsaye a matsakaicin tsayi, abin ya huta na ɗan lokaci (v _y = 0) kafin ya juya ƙasa. Saboda haka, saurin abu yana motsa parabolic a matsakaicin tsayi = v _x = v _ox

3. Idan abin ya motsa parabolic kamar yadda aka nuna a Hoto na 2, ana kallon bangaren tsaye na motsi parabolic a matsayin motsi na 'free-fall'. Idan abin ya motsa parabolic kamar Hoto na 1 da 3 to ana kallon bangaren tsaye na motsi parabolic a matsayin motsi na tsaye sama).

Misalan matsalolin:

1. Ana harba harsashi a kwance tare da saurin farko na 20 m/s. Idan bindigar ta kai mita 5 sama da ƙasa, a tantance:

(a) lokacin iska

(b) matsakaicin tsayi

(c) nisan kwance

(d) saurin harsashin lokacin da ya bugi ƙasa

Magani:

Ana nazarin motsi a cikin alkiblar kwance kamar motsi na layi ɗaya, yayin da ake nazarin motsi a cikin alkiblar tsaye kamar motsi na faɗuwa kyauta.

Wanda aka sani:

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

a) Lokacin iska

Maganin shine kamar ƙayyade tazara ta lokaci (t) a cikin motsi na faɗuwar kyauta.

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

Ana so: t

b) Matsakaicin tsayi

Tsawon da ya fi girma = h = mita 5.

c) Nisa a kwance (d)

Maganin kamar tantance nisan da ke kan motsi na layi ɗaya ne

An sani: v _ox = 20 m/s, t = daƙiƙa 1

Ana nema: d

d = vt

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

d) Sauri lokacin da harsashi ya faɗi ƙasa

v _tx = v _ox = 20 m/s

v _ty = ?

Da farko, muna ƙididdige gudun ƙarshe a tsaye (vty). Maganin kamar tantance saurin ƙarshe na motsi na 'free-fall' ne.

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

Ana so: v _ty

v _t = v _o + gt —> vo = 0

v _t = gt

v _t = (9.8 m/s ² )(1 s)

v _t = 9.8 m/s

Saurin harsashi idan ya faɗi ƙasa:

Hanyar harsashi:

Domin v _tx yana kan alkiblar x mai kyau (zuwa dama) kuma v _ty yana kan alkiblar y mara kyau (zuwa ƙasa),

alkiblar harsashin idan ya bugi ƙasa shine -26.1 ^o zuwa ga axis ɗin x mai kyau (duba hoton da ke ƙasa).

2. Bindigar ta harba harsashi a gudun 30 ^o zuwa kwance tare da gudun 60 m/s. Kayyade:

(a) matsakaicin tsayi

(b) saurin harsashi a matsakaicin tsayi

(c) lokacin iska

(d) nisan kwance

(e) saurin harsashin lokacin da ya bugi ƙasa. A ce ƙasa ba ta da faɗi. 🙂

Magani:

Ana nazarin motsi a cikin alkiblar kwance kamar motsi na layi ɗaya, motsi a cikin alkiblar tsaye ana nazarin motsi kamar motsi na tsaye sama.

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

Dangane da bayanai da aka sani, da farko za mu ƙididdige sassan tsaye (v _oy ) da kwance (v _ox ) na saurin farko (v _o ).

a) Matsakaicin tsayi (h)

Maganin kamar tantance matsakaicin tsayi ne akan motsi na tsaye sama.

Wanda aka sani:

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

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

a) Tsawo mafi girma (h)

Maganin kamar tantance matsakaicin tsayi ne a kan motsi na sama a tsaye.

Wanda aka sani:

v _oy = 30 m/s (wannan shine farkon saurin harsashin)

v _ty = 0 m/s (A matsakaicin tsayi, saurin tsaye na harsashi = 0 m/s. Wannan shine saurin ƙarshe.)

g = – 9.8 m/s ²

Ana so: h

v _t ² = v _o ² + 2 gh

0 ² = 30 ² + 2 (-9.8) h

0 = 900 – 19.6 hours

900 = awanni 19.6

h = 900/19.6

h = mita 45.9

Matsakaicin tsayin da harsashi ya samu = mita 45.9.

b) Gudun da ke kan matsakaicin tsayi

A matsakaicin tsayi, saurin da ke cikin alkiblar tsaye = 0 m/s. A matsakaicin tsayi, akwai kawai saurin da ke cikin alkiblar kwance. Saurin da ke cikin alkiblar kwance a matsakaicin tsayi daidai yake da saurin farko a alkiblar kwance, wanda shine 52.2 m/s. Alkiblar saurin da ke cikin alkiblar kwance koyaushe yana da daidaito, wato, a cikin alkiblar x-positive (idan an bayyana motsin abu a cikin zane da ke sama)

c) Lokacin iska

Maganin kamar tantance tazara ne (t) a cikin tattaunawar motsi na tsaye sama.

Wanda aka sani:

v _oy = 30 m/s (wannan shine saurin farko na harsashi a tsaye)

g = – 9.8 m/s ²

h = 0 m (lokacin da harsashin ya dawo ƙasa, juyawar harsashin a tsaye = 0 m)

Ana so: t

h = v _o t + 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 = 4.9t

t = 30 / 4.9

t = daƙiƙa 6.12

Lokacin da ake amfani da shi a iska = daƙiƙa 6.12

d) Nisa a kwance (d)

Maganin kamar tantance nisan (d) ne akan motsi na layi ɗaya.

Wanda aka sani:

v _ox = 52.2 m/s

t = daƙiƙa 6.12

Ana nema: d

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

e) Sauri lokacin da harsashi ya faɗi ƙasa

v _tx = v _ox = 52.2 m/s

v _ty = ?

Da farko, muna ƙididdige saurin ƙarshe a cikin alkiblar tsaye (v _ty ). Maganin shine a tantance saurin ƙarshe akan motsi na tsaye zuwa sama.

Ana so: v _oy = 30 m/s, g = -9.8 m/s ² , t = daƙiƙa 6.12

Ana so: v _ty

v _ty = v _oy + gt

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

v _ty = (30) – (60)

v _ty = -30 m/s

Alamar da ba ta da kyau tana nuna cewa alkiblar saurin ƙarshe tana ƙasa. Lura cewa saurin farko a alkiblar tsaye daidai yake da saurin ƙarshe a alkiblar tsaye.

Saurin harsashin idan ya bugi ƙasa:

Alkiblar harsashin:

Tunda v _tx yana kan alkiblar x mai kyau (zuwa dama) kuma vty yana kan alkiblar y mara kyau (zuwa ƙasa),

alkiblar saurin harsashin lokacin da ya bugi ƙasa shine -30 ^o game da axis mai kyau na x (duba hoton da ke ƙasa).

3. Ana jefa ƙwallon daga gefen ginin da ke da tsawon mita 50 tare da saurin farko na m 10/s. Idan an jefa ƙwallon a 30o kusa da kwance, a tantance:

(a) tazarar lokacin da ƙwallon ta kai ƙasa

(b) saurin ƙwallon lokacin da ya bugi ƙasa

(c) an auna nisan kwance da ƙwallon za ta iya isa da shi daga gefen ginin

(d) matsakaicin tsayin da ƙwallon ta kai

Magani:

Da farko, muna ƙididdige bangaren tsaye (v _oy ) da kuma ɓangaren kwance (v _ox ) na saurin farko (v _o ).

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

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

a) Tazarar lokaci ƙwallon ta isa ƙasa

Maganin kamar tantance tazara ta lokaci (t) ne a cikin motsi a tsaye sama. Girman vector wanda alkiblarsa ta sama an zaɓi shi ya zama mai kyau, girman vector wanda alkiblarsa ta ƙasa an zaɓi shi ya zama mara kyau. Matsayin ƙwallon da aka jefa shi an zaɓi shi a matsayin wurin tunani. h yana da korau saboda saman ƙasa yana ƙasa da wurin tunani, g yana da korau saboda alkiblar hanzarta nauyi tana ƙasa.

Wanda aka sani:

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

Ana so: t

h = v _o t + 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

Yi amfani da dabarar quadratic:

Lokacin da aka jefa ƙwallon a iska = tazara tsakanin lokacin da aka jefa ƙwallon zuwa ƙasa = daƙiƙa 1.64.

b) Saurin ƙwallon lokacin da ya bugi saman ƙasa

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

v _ty = ?

Da farko, muna ƙididdige gudun ƙarshe a tsaye (v _ty ). Maganin kamar tantance gudun ƙarshe ne a kan motsi na tsaye zuwa sama.

An sani: v _oy = 5 m/s, g = -9.8 m/s ² , t = daƙiƙa 1.64

Ana so: v _ty

v _ty = v _oy + gt

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

v _ty = 5 – 16

v _ty = -11 m/s

Alamar mara kyau tana nuna cewa alkiblar saurin ƙarshe tana ƙasa.

Saurin harsashin idan ya bugi ƙasa:

Alkiblar saurin harsashi = alkiblar motsin harsashin lokacin da ya bugi ƙasa:

Saurin harsashin idan ya bugi ƙasa:

Domin vtx yana kan alkiblar x mai kyau (zuwa dama) kuma vty yana kan alkiblar y mara kyau (zuwa ƙasa),

alkiblar harsashin idan ya bugi ƙasa shine -52o game da axis mai kyau na x (duba hoton da ke ƙasa).

c) Ana auna nisan kwance da ƙwallon za ta iya isa gare shi daga gefen ginin

Maganin kamar tantance nisan da aka yi tafiya a kai ne (d) a kan motsi na layi ɗaya.

Sananne: t = daƙiƙa 1.64, v _x = 8.7 m/s

Ana nema: d

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

d) Matsakaicin tsayin da ƙwallon ta kai

Sananne: v _oy = 5 m/s, v _ty = 0 m/s ( a tsaye ɓangaren gudun a matsakaicin tsayi = 0 m/s), g = -9.8 m/s ².

Ana so: 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 = mita 1.3

Tsawon da ƙwallon ya kai = mita 1.3 sama da saman ginin = mita 1.3 + mita 50 = mita 51.3 sama da saman ƙasa.