Misali na tambaya don tantance matsakaicin tsayin motsi na parabolic

2 Misalan tambayoyi don tantance matsakaicin tsayin motsi na parabolic

1. An harba ƙwallon sama a kusurwar 60^o a saman filin. Idan gudun farawa (v)_o) 10 m/s, menene matsakaicin tsayin ƙwallon? Saurin gudu saboda nauyi = 10 m/s²
Tattaunawa
An san cewa:
Kusurwoyi (θ) = 60^o
Saurin farko (v)_o) = 10 m/s
An tambaya: Tsawon mafi girma (h)
Amsa:

An kwatanta hanyar ƙwallon ta hanyar hoton.
Da farko ƙididdige saurin farko a tsaye:
ba 60^o = v_oy /v_o
v_oy = v_o ba 60^o = (10)(zunubi 60^o) = (10)(0,5√3) = 5√3 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:
Haɓaka nauyi (g) = -10 m/s² (mara kyau saboda alkiblar saurin nauyi tana ƙasa)
Saurin farko a tsaye (v)_oy) = +5√3 m/s (mai kyau saboda alkiblar saurin 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.
An tambaya: Tsawon mafi girma (h)
Amsa:
Domin kuwa adadin da aka sani shine v_oy, g da v_ty, yayin da tambayar ita ce h, to dabarar motsi na tsaye sama da aka yi amfani da ita ita ce:
v_t² = v_o² + 2 gh
Bayani: v_t = gudun ƙarshe, v_o = saurin farko, g = hanzari saboda nauyi, h = matsakaicin tsayi.

Tsawon mafi girma:
v_t² = v_o² + 2 gh
02 = (5√3)² + 2 (-10) h
0 = 25(3) – awanni 20
0 = 75 – 20 hours
75 = awanni 20
h = 75/20
h = mita 3,75
Don haka matsakaicin tsayin da ƙwallon ta kai shine mita 3,75.

2. Ana jefa wani abu sama a kusurwar 30°^o a kwance, daga ginin bene mai hawa da yawa mita 20 sama da ƙasa. Saurin farko na abin shine m4/s. Kayyade matsakaicin tsayin abin daga ƙasa! Saurin hanzarin da aka samu sakamakon nauyi shine m10/s².
Tattaunawa
An san cewa:
Kusurwoyi (θ) = 30^o
Tsawon gini (h) = mita 20
Saurin farko (v)_o) = 4 m/s
Saurin gudu saboda nauyi (g) = 10 m/s²
An tambaya: An ƙididdige matsakaicin tsayin abin daga saman ƙasa (h)
Amsa:
Saurin farko na abu a tsaye:
ba 30^o = v_oy /v_o
v_oy = v_o ba 30^o = (4)(zunubi 30^o) = (4)(0,5) = 2 m/s

Yanzu ƙididdige matsakaicin tsayin da abin ya kai daga ginin bene mai hawa da yawa ta amfani da dabarar motsi na tsaye sama. Sannan, ƙara wannan zuwa tsayin ginin don samun jimlar tsayin. Lokacin magance matsalolin motsi na tsaye sama, ana ba da adadin vector da ke nuna sama alama mai kyau, kuma adadin vector da ke nuna ƙasa alama mara kyau.

An san cewa:
Saurin gudu saboda nauyi (g) = -10 m/s² (mara kyau saboda alkiblar saurin nauyi tana ƙasa)
Saurin farko a tsaye (v)_oy) = +2 m/s (mai kyau saboda alkiblar saurin yana sama)
Gudun abu a matsakaicin tsayi (v)_ty) = 0
An tambaya: Tsawon mafi girma (h)
Amsa:
Tsawon mafi girma:
v_t² = v_o² + 2 gh
0² = 2² + 2 (-10) h
0 = 4 – 20 hours
4 = awanni 20
h = 4/20
h = mita 0,2
Matsakaicin tsayin da ƙwallon da aka ƙididdige daga saman ƙasa ya kai shine mita 0,2 + mita 20 = mita 20,2.

[Turanci: Magance matsalolin motsin harsashi - tantance matsakaicin tsayin daka ]