Misali na motsi na parabolic

Misalai 7 na Tambayoyin Motsin Parabolic

1. Ana harba harsashi a gudun 20 ms ^–1 . Idan kusurwar ɗagawa ita ce 60 ^o kuma hanzarin da aka samu saboda nauyi = 10 ms ^–2 , to harsashin ya kai matsayi mafi girma bayan…

A. Daƙiƙa 1

B. Daƙiƙa 2

C. √Daƙiƙa 3

D. 2√Daƙiƙa 3

E. 3√2 daƙiƙa

Tattaunawa

An san cewa:

Saurin farko na harsashi (v _o ) = 20 ms ^–1

Kusurwar tsayi (θ) = 60 ^o C

Saurin gudu saboda nauyi (g) = 10 ms ^–2

Tambaya: Tazarar lokacin da harsashi zai kai ga mafi girman matsayi

Amsa:

Saurin farko na harsashi a cikin alkiblar kwance (axis-x):

v _ox = v _o cos 60 ^o = (20) (0,5) = 10 m/s

Saurin farko na harsashi a tsaye (axis-y):

v _oy = v _o zunubi 60 ^o = (20)(0,5√3) = 10√3 m/s

Domin ƙididdige tazarar lokacin da harsashi zai kai ga matsakaicin tsayinsa, a duba motsin harsashin tun daga lokacin da aka harba shi har sai ya kai matsakaicin tsayinsa. A mafi girman wurinsa, harsashin ya tsaya na ɗan lokaci kafin ya juya alkibla, don haka saurinsa a mafi girman wurinsa sifili ne (v _ty = 0).

Ana ƙididdige tazarar lokacin da harsashi zai kai ga mafi girman matsayi ta amfani da dabarar da ke ƙasa:

v _ty = v _oy + gt

Bayani:

v _ty = gudun ƙarshe na harsashi a tsaye = gudun harsashi a mafi girman matsayi = 0 m/s

v _oy = saurin farko na harsashi a tsaye = 10√3 m/s

g = hanzari saboda nauyi = 10 m/s ²

t = tazara tsakanin lokaci

Tazarar lokaci don harsashi ya kai ga mafi girman matsayi:

v _ty = v _oy + gt

0 = 10√3 – 10 t

10√3 = t 10

t = 10√3 / 10

t = √ daƙiƙa 3

Amsar da ta dace ita ce C.

2. Ana harba harsashi da saurin V _o da kusurwar α. A mafi girman matsayi, to...

A. kuzarin motsi sifili ne

B. matsakaicin kuzarin motsi

C. matsakaicin makamashi mai yuwuwa

D. jimlar ƙarfin shine mafi girma

E. matsakaicin gudu

Tattaunawa

Idan aka harba harsashi da saurin farko v _o da kusurwar ɗagawa α, harsashin yana motsawa a cikin parabola. A matsakaicin tsayi, ƙarfin ƙarfin nauyi yana kan iyakarsa saboda harsashin yana kan matsakaicin tsayinsa. A mafi girman matsayi, harsashin yana ci gaba da motsawa a kwance saboda harsashin yana da ƙarfin motsi, kodayake ƙimarsa ba ta da yawa. Ƙarfin motsi yana kan mafi ƙarancinsa saboda yawancin kuzarin ana canza shi zuwa ƙarfin motsi.

Amsar da ta dace ita ce C.

3. Mai tsaron gida yana buga ƙwallon da hanyar da aka nuna a hoton. Nisa X shine…. (g = 10 ms ^-2 ).

A. 62,5 m

B. 31,25 √ 2 m

C. mita 31,25

D. 25 √ 2 m

E. 25 m

Tattaunawa

An san cewa:

Saurin farko (v _o ) = 25 m/s

Saurin gudu saboda nauyi (g) = 10 m/s ²

Kusurwoyi (θ) = 45 ^o

An tambaya: Distance X

Amsa:

Saurin farko na ƙwallon a cikin alkiblar kwance:

v _ox = v _o cos θ = (25 m/s) (cos 45 ^o ) = (25 m/s) (0,5 √ 2 ) = 12,5 √ 2 m/s

Saurin farko na ƙwallon a tsaye:

v _oy = v _o sin θ = (25 m/s) (sin 45 ^o ) = (25 m/s) (0,5 √ 2) = 12,5 √ 2 m/s

Motsin Parabolic haɗuwa ce ta motsi a kwance da a tsaye. Saboda haka, ana nazarin motsin parabolic kamar an haɗa shi da motsi biyu daban-daban. Ana nazarin motsin kwance a matsayin motsi na layi ɗaya , kuma ana nazarin motsi na tsaye a matsayin motsi na tsaye sama.

Tazarar lokaci na ƙwallon a cikin iska (t):

Da farko, ƙididdige tazarar lokacin da ƙwallon za ta motsa tare da parabola. Ana ƙididdige tazarar lokacin ta amfani da dabarar motsi sama a tsaye.

A wajen magance matsaloli a kan motsi a tsaye sama, adadin vektor da aka nuna sama ana ba shi alama mai kyau, adadin vektor da aka nuna ƙasa ana ba shi alama mara kyau.

An san cewa:

Saurin farko (v _o ) = 12,5 √ 2 m/s (mai kyau saboda alkiblar saurin farko tana sama)

Saurin gudu saboda nauyi (g) = -10 m/s ² (mara kyau saboda alkiblar hanzari saboda nauyi yana ƙasa)

Tsawo (h) = 0 (lokacin da ƙwallon ta koma matsayinta na asali, canjin tsayin ƙwallon sifili ne)

Tambaya: Tazarar lokaci (t) da ƙwallon ke motsawa a kan parabola

Amsa:

An ba da v _o , g, h kuma an nemi t don haka dabarar motsi a tsaye sama da aka yi amfani da ita ita ce h = v _o t + 1/2 gt ^{​​​​2}

h = v _o t + 1/2 gt ^{​​​​2}

0 = ( 12,5 √ 2 ) t + 1/2 (-10) t ²

0 = 12,5 √ 2 t – 5 t ²

12,5 √ 2 t = 5 t ²

12,5 √ 2 = 5 t

t = 12,5 √ 2/5

t = 2,5 √ Daƙiƙa 2

Nisa ta kwance da ƙwallon ta kai (X):

Ana ƙididdige nisan kwance ta amfani da dabarar motsi mai layi ɗaya.

An san cewa:

Gudun (v) = 12,5 √ 2 m/s

Tazarar lokaci (t) = 2,5 √ daƙiƙa 2

An tambaya: Nisa

Amsa:

s = vt = ( 12,5 √ 2 )( 2,5 √ 2 ) = ( 12,5 )( 2,5 )(2) = mita 62,5

Amsar da ta dace ita ce A.

4. An harba harsashin da wata hanya kamar yadda aka nuna a hoton (g = 10 ms ^-2 )

Tsawon da harsashin ya kai shine...

A. 5 m

B. 10 m

C. 20 mita

D. 25 m

E. mita 30

Tattaunawa

An san cewa:

Saurin farko (v _o ) = 20 m/s

Saurin gudu saboda nauyi (g) = 10 m/s ²

Kusurwoyi (θ) = 30 ^o

An tambaya: Tsawon da ya fi tsayi (h mafi girma)

Amsa:

Da farko, ƙididdige saurin farko a tsaye (v _oy ):

v _oy = v _o zunubi 30 ^o = (20)(zunubi 30 ^o ) = (20) (0,5) = 10 m/s

Bayan samun ƙimar saurin farko a cikin alkiblar tsaye (v _oy ) , yanzu ƙididdige matsakaicin tsayi ta amfani da hanya ɗaya da lissafin matsakaicin tsayi a cikin motsi na tsaye sama . Wajen magance matsalolin motsi na tsaye sama, adadin vector da aka nuna sama ana ba shi alama mai kyau, adadin vector da aka nuna ƙasa ana ba shi alama mara kyau.

An san cewa:

Saurin gudu saboda nauyi (g) = -10 m/s ² (mara kyau saboda alkiblar hanzari saboda nauyi yana ƙasa)

Gudun farko a tsaye (v _oy ) = 10 m/s (positive saboda alkiblar gudun yana sama)

Gudun da ke da matsakaicin tsayi (v _ty ) = 0

A matsakaicin tsayi, abu yana hutawa na ɗan lokaci kafin ya koma ƙasa. Don haka a matsakaicin tsayi, saurin abu sifili ne.

Ana so: Matsakaicin tsayi (h)

Amsa:

Saboda yawan da aka sani sune v _oy , g da v _ty , yayin da wanda ake tambaya shine h, dabarar motsi na tsaye sama da aka yi amfani da ita ita ce:

v _t ² = v _o ² + 2 gh

Bayani: v _t = gudun ƙarshe, v _o = gudun farko, g = hanzarin nauyi, h = matsakaicin tsayi.

Tsawon mafi girma:

v _t ² = v _o ² + 2 gh

0 ² = 10 ² + 2 (-10) h

0 = 100 – 20 hours

100 = awanni 20

h = 100/20

h = mita 5

Tsawon mafi girma shine mita 5.

Amsar da ta dace ita ce A.

5. Mutum yana riƙe da ƙwallo a tsayin mita 20 sannan ya jefa ta a kwance gaba da saurin farko na mita 5/s. A ƙayyade:
(a) Tazarar lokacin da ƙwallon za ta isa ƙasa
(b) Mafi girman nisan kwance da ƙwallon ta kai
(c) Saurin ƙwallon idan ya buga ƙasa

Tattaunawa

(a) Tazarar lokacin da ƙwallon zai isa ƙasa (t)
Maganin kamar tantance tazara ne ga wani abu da ke cikin faɗuwar free fall.

(b) Mafi girman nisan kwance da ƙwallon (s) ta kai

An san cewa:
v_ox = 5 m/s (gudun farko a kwance)
t = daƙiƙa 2 (tazara tsakanin lokacin ƙwallon a cikin iska)
An tambaya: s
Amsa:
v = s / t
s = vt = (5)(2) = mita 10
(c) Saurin ƙwallon idan ya buga ƙasa (v)_t)
v_ox = v_tx = v_x = 5m/s
v_ty = ….?
Ana ƙididdige gudu na ƙarshe a tsaye kamar ana ƙididdige gudu na ƙarshe a cikin motsi na faɗuwa kyauta.
An sani: v_oy = 0, g = 10, h = 20
An tambaya: v_t
Amsa:

6. An buga ƙwallon a kusurwar 30^o a saman filin da saurin farko na m10/s. A ƙayyade:
(a) Matsakaicin tsayi
(b) Saurin ƙwallon a matsakaicin tsayi
(c) Tazarar lokacin da ƙwallon za ta isa saman filin
(d) Mafi girman nisan kwance da ƙwallon ta kai

Tattaunawa

(a) Matsakaicin tsayi
Maganin kamar tantance matsakaicin tsayi ne a cikin motsi na tsaye sama.
An san cewa:
v_o = 10m/s
v_oy = v_o zunubi 30 = (10)(0,5) = 5 m/s
g = -10 m/s²
v_ty = 0
An tambaya: matsakaicin h
(b) Saurin ƙwallon a matsakaicin tsayi
Gudun a matsakaicin tsayi = gudun a kwance = v_x.
v_x = v_o cos 30 = (10)(0,87) = 8,7 m/s
(c) Tazarar lokaci
Maganin kamar tantance tazara ne na lokacin motsi a tsaye sama.
An san cewa:
v_oy = v_o zunubi 30 = (10)(0,5) = 5 m/s
g = -10 m/s²
da h = 0
An tambaya: t
Amsa:
(d) Nisa mafi nisa a kwance
x = v_x t = (8,7)(1) = mita 8,7

7. Ana jefa ƙwallon daga gefen ginin da ke da tsawon mita 10, wanda ke samar da kusurwar 30°.^o zuwa kwance tare da saurin farko na 10 m/s.
(a) Matsakaicin tsayi da aka auna daga matakin ƙasa
(b) Tazarar lokacin da ƙwallon za ta isa ƙasa
(c) nisan kwance mafi nisa da aka auna daga gefen ginin

Tattaunawa
(a) Matsakaicin tsayi da aka auna daga matakin ƙasa
Maganin kamar tantance matsakaicin tsayi ne a cikin motsi na tsaye sama.
A ƙididdige tsayin ƙwallon da aka auna daga gefen ginin da aka jefa ƙwallon daga ciki.. Duba motsin ƙwallon tun daga lokacin da aka jefa ta har sai ta kai ga matsakaicin tsayinta.
An san cewa:
v_o = 10m/s
v_oy = v_o ba 30^o = (10)(0,5) = 5 m/s
v_ty = 0 (a matsakaicin tsayi, abu yana hutawa na ɗan lokaci)
g = -10 m/s²
An tambaya: h

(b) Tazarar lokacin da ƙwallon za ta isa ƙasa
Maganin yana kama da tantance tazarar lokacin motsi a tsaye sama. Yi la'akari da motsin ƙwallon tun daga lokacin da aka jefa shi har sai ya isa ƙasa.
An san cewa:
v_o = 10m/s
v_oy = v_o ba 30^o = (10)(0,5) = 5 m/s
g = -10 m/s²
h = -10 m (matsayin ƙarshe shine mita 10 ƙasa da matsayin farko)
An tambaya: t

Ba zai yiwu lokaci ya sami ƙimar da ba ta da kyau ba, saboda haka t = daƙiƙa 2.
(c) An auna nisan kwance mafi nisa daga gefen ginin
v_o = 10m/s
v_x = v_ox = v_o cos 30 = (10)(0,87) = 8,7 m/s
t = daƙiƙa 2
Nisa mafi nisa a kwance:

s = v _x t = (8,7)(2) = mita 17,4

Tambayoyi game da motsi na parabolic / motsi na harsashi

1. Mutum yana riƙe da ƙwallo a tsayin mita 5 sannan ya jefa ta a kwance gaba da saurin farko na mita 2/s. A ƙayyade:
(a) Tazarar lokacin da ƙwallon za ta isa ƙasa
(b) Mafi girman nisan kwance da ƙwallon ta kai
(c) Saurin harsashin idan ya bugi ƙasa
Yi amfani da g = 10 m/s²
Amsa:
(a) t = 1 s
(b) s = 2 m
(c) v_t = 10,2m/s

2. An buga ƙwallon a kusurwar 60^o a saman filin da saurin farko na m5/s. A ƙayyade:
(a) Matsakaicin tsayi
(b) Saurin ƙwallon a matsakaicin tsayi
(c) Tazarar lokacin da ƙwallon za ta isa saman filin
(d) Mafi girman nisan kwance da ƙwallon ta kai
Yi amfani da g = 10 m/s²
Amsa:
(a) h = 1 m (mai zagaye)
(b) v = v_x = 2,5m/s
(c) t = 0,87 s
(d) x = 2,175 m
3. Ana jefa ƙwallon daga gefen ginin da ke da tsawon mita 5, wanda ke samar da kusurwar 60°.^o zuwa kwance tare da saurin farko na 5 m/s.
(a) Matsakaicin tsayi da aka auna daga matakin ƙasa
(b) Tazarar lokacin da ƙwallon za ta isa ƙasa
(c) An auna nisan kwance mafi nisa daga gefen ginin
Yi amfani da g = 10 m/s²
Amsa:
(a) h = 5,95 m
(b) t = s 1,5
(c) x = 3,75 m

Tushen tambaya:

Tambayoyin Nazarin Fizik na Ƙasa ga Makarantar Sakandare ta Babbar Sakandare/Makarantar Sakandare ta Sana'a